1. The figure shows a 100-kg block being released from rest from a height of 1.0 m. It then takes 0.53 s for it to reach the floor. What is the mass m of the block on the left? There is no friction or mass in the pulley, and the connecting rope is very light.
A) 16 kg B) 14 kg C) 13 kg D) 11 kg
2. As shown in the figure, a 10-kg block on a perfectly smooth horizontal table is connected by a horizontal string to a 63-kg block that is hanging over the edge of the table. What is the magnitude of the acceleration of the 10-kg block when the other block is gently released?
A) 8.5 m/s2 B) 8.1 m/s2 C) 7.5 m/s2 D) 9.0 m/s2
3. A 958-N rocket is coming in for a vertical landing. It starts with a downward speed of 25 m/s and must reduce its speed to 0 m/s in 8.0 s for the final landing.
(a) During this landing maneuver, what must be the thrust due to the rocket's engines?
(b) What must be the direction of the engine thrust force?
4. A 200-N sled of slides down a frictionless hillside that rises at 37° above the horizontal. What is the magnitude of the force that the hill exerts on the sled parallel to the surface of the hill?
A) 200 N B) 120 N C) 160 N D) 150 N E) 0 N
5. The coefficients of static and kinetic friction between a 3.0-kg box and a horizontal desktop are 0.40 and 0.30, respectively. What is the force of friction on the box when a 15-N horizontal push is applied to the box?
A) 12 N B) 8.8 N C) 15 N D) 6.0 N E) 4.5 N
6. A 10-kg sign is held by two ropes as shown in the figure. What is the tension on rope A?
A) 44 N B) 69 N C) 72 N D) 88 N E) 98 N
7. A 50-kg box is being pushed along a horizontal surface. The coefficient of static friction between the box and the ground is 0.65, and the coefficient of kinetic friction is 0.35. What horizontal force must be exerted on the box for it to accelerate at 1.2 m/s2?
A) 60 N B) 120 N C) 170 N D) 230 N E) 490 N
8. As shown in the figure, block B on a horizontal tabletop is attached by very light horizontal strings to two hanging blocks, A and C. The pulleys are ideal, and the coefficient of kinetic friction between block B and the tabletop is 0.100. The masses of the three blocks are mA = 12.0 kg, mB = 7.00 kg, and mC = 10.0 kg. Find the magnitude and direction of the acceleration of block B after the system is gently released and has begun to move.
9. The figure shows a block of mass m resting on a 20° slope. The block has coefficients of friction and with the surface of the slope. It is connected using a very light string over an ideal pulley to a hanging block of mass 2.0 kg. The string above the slope pulls parallel to the surface. What is the minimum mass m so the system will remain at rest when it is released from rest?
A) 2.1 kg B) 3.3 kg C) 1.3 kg D) 3.6 kg
Check my answers?
1.C, 2.B, 3.?, 4.E, 5.D, 6.A(angle A=45, B=30, 7.D, 8. To the left,? (block A is hanging off the left side, block C is on the right), 9.?
To check the answers for the given questions, let's go through each question and explain how to solve them.
1. The figure shows a 100-kg block being released from rest from a height of 1.0 m. It then takes 0.53 s for it to reach the floor. What is the mass m of the block on the left? There is no friction or mass in the pulley, and the connecting rope is very light.
To solve this question, we can use the principle of conservation of energy. The potential energy of the block at the height of 1.0 m is given by mgh, where m is the mass, g is the acceleration due to gravity, and h is the height. The kinetic energy at the bottom is given by 1/2mv^2, where v is the velocity. Since the block is released from rest, the initial velocity is 0.
Assuming no energy losses, the potential energy at the top is equal to the kinetic energy at the bottom. Therefore, we can write the equation as:
mgh = 1/2mv^2
We can cancel out the mass and solve for the velocity:
v = sqrt(2gh)
Substituting the given values of g = 9.8 m/s^2 and h = 1.0 m, we get:
v = sqrt(2 * 9.8 * 1.0) = sqrt(19.6) ≈ 4.43 m/s
Now, we can use the equation for average velocity:
v = d/t
Rearranging the equation for t, we get:
t = d/v
Substituting the given values of v = 4.43 m/s and d = 0.53 m, we get:
t = 0.53 s
Since the block travels a distance of 1.0 m in 0.53 s, the velocity is constant, and we can use the equation of motion:
v = d/t
Substituting the given values, we get:
v = 1.0 m / 0.53 s ≈ 1.89 m/s
Now, we can use the equation for momentum, which is given by:
p = mv
Substituting the given value of v = 1.89 m/s, we get:
p = m * 1.89
We know that momentum is conserved, so the momentum of the block on the left is equal to the momentum of the falling block:
p = 100 * 1.89 = m * 4.43
Solving for m, we get:
m = (100 * 1.89) / 4.43 ≈ 42.72 kg
Therefore, the mass m of the block on the left is approximately 42.72 kg.
The correct answer is not listed in the options provided, so none of the options (A, B, C, D) are correct.
2. As shown in the figure, a 10-kg block on a perfectly smooth horizontal table is connected by a horizontal string to a 63-kg block that is hanging over the edge of the table. What is the magnitude of the acceleration of the 10-kg block when the other block is gently released?
In this scenario, we can assume that the only force acting on the 10-kg block is the tension in the string. This tension force is equal in magnitude to the weight of the hanging block (63 kg), which is given by:
T = m * g = 63 * 9.8 = 617.4 N
Since there is no friction or external horizontal force, the acceleration of the system will be equal for both blocks. We can use Newton's second law, which states that the net force on an object is equal to the product of mass and acceleration:
T - m * g = m * a
Substituting the values we know, we get:
617.4 - (10 * 9.8) = 10 * a
a = (617.4 - 98) / 10 = 519.4 / 10 ≈ 51.94 m/s^2
Therefore, the magnitude of the acceleration of the 10-kg block when the other block is gently released is approximately 51.94 m/s^2.
The correct answer is not listed in the options provided, so none of the options (A, B, C, D) are correct.
1. To find the mass of the block on the left (m), we can use the principle of conservation of energy.
Potential energy = Kinetic energy
mgh = (1/2)mv^2
where m is the mass, g is the acceleration due to gravity, h is the height, and v is the final velocity.
Rearranging the equation, we get:
m = (1/2)(v^2)/g
We can calculate the final velocity using the equation:
v = gt
where t is the time taken to reach the floor.
Substituting the values, we have:
v = (9.8 m/s^2)(0.53 s) = 5.194 m/s
Now, substituting the values into the mass equation, we get:
m = (1/2)(5.194 m/s)^2 / (9.8 m/s^2) = 1.37 kg
Therefore, the mass of the block on the left is approximately 1.37 kg. The closest option is C) 13 kg.
2. To find the magnitude of acceleration of the 10-kg block, we can use Newton's second law of motion:
Force = Mass x Acceleration
The force acting on the 10-kg block is the tension in the string. The force exerted by the hanging block is its weight, which is mg, where m is the mass and g is the acceleration due to gravity.
The equation can be written as:
T - mg = ma
where T is the tension, m is the mass of the 10-kg block, g is the acceleration due to gravity, and a is the acceleration of the block.
Since the 10-kg block is on a smooth surface, there is no friction, and the tension T is the only external force acting on it.
Rearranging the equation, we have:
a = (T - mg) / m
Substituting the known values, we get:
a = (T - 10 kg * 9.8 m/s^2) / 10 kg
Now, the tension T can be found by considering the hanging block. The hanging block experiences a force equal to its weight, mg, which is balanced by the tension T. Therefore, T = mg.
Substituting this value, we get:
a = (mg - mg) / m = 0
Therefore, the magnitude of the acceleration of the 10-kg block is 0 m/s^2. The closest option is D) 9.0 m/s^2.
3. (a) To determine the thrust due to the rocket's engines, we can use Newton's second law of motion.
Force = Mass x Acceleration
The net force acting on the rocket is equal to its mass multiplied by its acceleration. Since the rocket is coming in for a vertical landing, we can assume that the acceleration is in the opposite direction to its initial velocity (upwards).
Therefore, the net force acting on the rocket is:
Thrust - Weight = mass x acceleration
Since the final velocity is 0 m/s, the acceleration is equal to the change in velocity divided by the time taken:
acceleration = (0 m/s - (-25 m/s)) / 8.0 s = 3.125 m/s^2
The weight of the rocket can be found using the equation:
Weight = mass x gravitational acceleration
Weight = 958 N = mass x (9.8 m/s^2)
Rearranging the equation, we get:
mass = 958 N / 9.8 m/s^2 ≈ 97.96 kg
Now, substituting the known values into the net force equation, we have:
Thrust - 97.96 kg x 3.125 m/s^2 = 0
Thrust = 97.96 kg x 3.125 m/s^2 ≈ 306.13 N
Therefore, the thrust due to the rocket's engines is approximately 306.13 N.
(b) The direction of the engine thrust force is opposite to the initial velocity of the rocket, which is downwards (since the rocket is coming in for a vertical landing).