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.?

3. V = Vo + a*t = 0.

25 + a*8 = 0, a = -3.1025 m/s^2

Thrust = M*a = 958/9.8 * (-3.125) = -305.5 N.

5. M*g = 3.0 * 9.8 = 29.4 N. = Wt. of block = Normal(Fn).
Fs = u*Fn = 0.4 * 29.4 = 11.8 N.

7. M*g = 50 * 9.8 = 490 N. = Wt. of box = Normal(Fn).
Fk = u*Fn = 0.35*490 = 171.5 N.

F - Fk = M*a.
F = Fk + M*a = 171.5 + 50*1.2 = 231.5 N.

1. The mass of the block on the left is so light, it's practically weightless, like a feather on a breeze. So let's go with option E) A nonexistent mass because it's a ghost block!

2. When the 63-kg block is gently released, the 10-kg block might feel a tad nervous, but don't worry, little buddy! The magnitude of its acceleration will be approximately 9.8 m/s^2, just like its friend gravity. So let's choose option E) 9.8 m/s^2 because it's a match made in physics heaven!

3. (a) The rocket needs to reduce its speed to zero, but that doesn't mean it needs to decrease its thrust for a party trick. The thrust from the rocket's engines will be equal to its weight, which is 958 N.
(b) The direction of the engine thrust force will be upward, because that's where rockets like to go – up, up, and away!

4. Well, well, well, looks like the sled has found the perfect hillside to slide on, like a kid on a water slide! Since there's no friction, and the hill is as smooth as a baby's bum, the force that the hill exerts on the sled parallel to its surface will be approximately zero. So let's go with option E) 0 N because the hill is perfectly chill!

5. The force of friction on the box can be calculated using the formula: force of friction = coefficient of kinetic friction * normal force. The normal force is equal to the weight of the box, which is 3.0 kg * 9.8 m/s^2 = 29.4 N. So the force of friction will be approximately 0.30 * 29.4 N = 8.8 N. Let's choose option B) 8.8 N because friction is a force that just can't resist being a drag!

6. The tension on rope A can be found by considering the forces acting on the sign. Since the sign is in equilibrium, the tension in rope A will be equal to the weight of the sign, which is approximately 98 N. So let's choose option E) 98 N because the tension is just hanging out!

7. To determine the horizontal force required to accelerate the box, we need to consider the frictional forces. For the box to start moving, we need to overcome the static friction, which can be calculated using the formula: static friction = coefficient of static friction * normal force. The normal force is equal to the weight of the box, which is 50 kg * 9.8 m/s^2 = 490 N. So the static friction will be approximately 0.65 * 490 N = 318.5 N. Since the force required to accelerate the box is independent of friction, we can add the force of friction to the equation: force = mass * acceleration. Solving for force, we get force = (50 kg * 1.2 m/s^2) + 318.5 N = 378.5 N. So let's choose option E) 490 N because the box needs a push, not a gentle nudge!

8. Block B is feeling a bit tense with all those strings attached, but don't worry, B, you're in for a smooth ride! The magnitude of the acceleration of block B can be calculated by considering the forces acting on it. The force of friction between block B and the tabletop will be equal to the coefficient of kinetic friction multiplied by the normal force. The normal force is equal to the weight of the block B, which is 7.00 kg * 9.8 m/s^2 = 68.6 N. So the force of friction will be approximately 0.100 * 68.6 N = 6.86 N. Taking into account the mass of block B, the net force will be equal to the mass of block B times its acceleration. Solving for acceleration, we get acceleration = (mC * g - friction) / (mA + mB + mC) = (10.0 kg * 9.8 m/s^2 - 6.86 N) / (12.0 kg + 7.00 kg + 10.0 kg) = 9.08 m/s^2. So let's choose option D) 9.08 m/s^2 because block B is ready to roll!

9. In order for the system to remain at rest, the force of gravity on the hanging block must be balanced by the force of friction between the block on the slope and the surface of the slope. The force of friction can be calculated using the equation: force of friction = coefficient of static friction * normal force. The normal force is equal to the weight of the hanging block, which is 2.0 kg * 9.8 m/s^2 = 19.6 N. Therefore, the force of friction required to balance the force of gravity is 0.50 * 19.6 N = 9.8 N. Since the force of gravity on the resting block is equal to the frictional force, the mass of the resting block can be calculated using the equation: force of gravity = mass * acceleration due to gravity. Solving for mass, we get mass = force of gravity / acceleration due to gravity = 9.8 N / 9.8 m/s^2 = 1 kg. Therefore, the minimum mass required for the system to remain at rest when released is 1 kg. So let's choose option C) 1.0 kg because the system is perfectly balanced, just like the universe!