Part 1
A 3.0 kg block is pushed 3.0 m at a constant
velocity up a vertical wall by a constant force
applied at an angle of 30.0
◦
with the horizontal, as shown in the figure.
The acceleration of gravity is 9.81 m/s2.
If the coefficient of kinetic friction between
the block and the wall is 0.30, find
a) the work done by the force on the block.
Answer in units of J.
Part 2
b) the work done by gravity on the block.
Answer in units of J.
Part 3
c) the magnitude of the normal force between
the block and the wall.
Answer in units of N.
what does that mean Henry?? We trying to cheat, not learn the whole lesson homie.
Part 1: To find the work done by the force on the block, we need to calculate the component of the force parallel to the displacement and then multiply it by the displacement.
The force applied at an angle of 30.0° with the horizontal can be split into two components: the vertical component and the horizontal component.
Vertical component = force * sin(angle) = F * sin(30°)
Horizontal component = force * cos(angle) = F * cos(30°)
Since the block is being pushed vertically, the vertical component of the force is contributing to the work done.
Work done = vertical component of force * displacement = F * sin(30°) * displacement
Now, let's calculate the work done:
Work done = (mass * gravity) * sin(30°) * displacement
= (3.0 kg * 9.81 m/s^2) * sin(30°) * 3.0 m
= 44.14 J
Therefore, the work done by the force on the block is 44.14 J.
Part 2: The work done by gravity on the block can be calculated by multiplying the weight of the block by the vertical displacement.
Weight of the block = mass * gravity
Work done by gravity = weight of the block * displacement
Now, let's calculate the work done by gravity:
Work done by gravity = (mass * gravity) * displacement
= (3.0 kg * 9.81 m/s^2) * 3.0 m
= 88.29 J
Therefore, the work done by gravity on the block is 88.29 J.
Part 3: The normal force between the block and the wall is equal in magnitude and opposite in direction to the gravitational force acting on the block. Therefore, it is equal to the weight of the block.
Normal force = weight of the block = mass * gravity
Normal force = 3.0 kg * 9.81 m/s^2
= 29.43 N
Therefore, the magnitude of the normal force between the block and the wall is 29.43 N.
Part 1:
To find the work done by the force on the block, we can use the formula:
Work = Force * Distance * cos(angle)
In this case, the force is the applied force pushing the block up the wall, and the distance is the distance the block is pushed. The angle between the force and the horizontal is given as 30 degrees.
Given:
Force = ?
Distance = 3.0 m
Angle = 30 degrees
First, we need to find the force applied. Since the block is moving at a constant velocity, the net force acting on it must be zero.
The net force can be calculated using the equation:
Net Force = Applied Force - Friction Force - Weight
The friction force can be calculated using the formula:
Friction Force = Coefficient of Friction * Normal Force
The weight of the block can be calculated using the formula:
Weight = mass * gravity
Given:
Mass = 3.0 kg
Coefficient of Friction = 0.30
Gravity = 9.81 m/s^2
Now we can substitute the known values into the formulas to find the force applied, friction force, and weight.
Weight = mass * gravity
Weight = 3.0 kg * 9.81 m/s^2
Weight = 29.43 N
Friction Force = Coefficient of Friction * Normal Force
Friction Force = 0.30 * Normal Force
Since the block is moving up the wall at a constant velocity, the normal force must be equal to the weight of the block.
Normal Force = Weight
Normal Force = 29.43 N
Now we can find the friction force:
Friction Force = 0.30 * Normal Force
Friction Force = 0.30 * 29.43 N
Friction Force = 8.829 N
Now we can substitute the values into the net force equation:
Net Force = Applied Force - Friction Force - Weight
0 = Applied Force - 8.829 N - 29.43 N
Applied Force = 38.259 N
Now we can calculate the work done by the force:
Work = Force * Distance * cos(angle)
Work = 38.259 N * 3.0 m * cos(30 degrees)
Work = 99.95 J
Therefore, the work done by the force on the block is 99.95 Joules (J).
Part 2:
To find the work done by gravity on the block, we can use the formula:
Work = Force * Distance * cos(angle)
In this case, the force is the weight of the block, and the distance is the vertical distance it has been moved.
Given:
Force = Weight = 29.43 N
Distance = 3.0 m
Angle = 180 degrees (as gravity acts vertically downward)
Now we can calculate the work done by gravity:
Work = Force * Distance * cos(angle)
Work = 29.43 N * 3.0 m * cos(180 degrees)
Work = -88.29 J
Therefore, the work done by gravity on the block is -88.29 Joules (J). The negative sign indicates that the work done by gravity is in the opposite direction to the applied force.
Part 3:
The magnitude of the normal force between the block and the wall is equal to the weight of the block.
Normal Force = Weight
Normal Force = 29.43 N
Therefore, the magnitude of the normal force between the block and the wall is 29.43 Newtons (N).
Part 1:
To find the work done by the force on the block, use the formula for work: W = F * d * cos(theta), where F is the force, d is the displacement, and theta is the angle between the force and the displacement.
In this case, the force is the component of the applied force in the direction of the displacement, which is F = F_applied * cos(theta). The displacement is given as 3.0 m. The angle theta is 30.0 degrees.
Step 1: Calculate the force component in the direction of the displacement:
F = F_applied * cos(theta)
F = F_applied * cos(30.0 degrees)
F = F_applied * √3/2
Step 2: Calculate the work done:
W = F * d * cos(theta)
W = (F_applied * √3/2) * 3.0 m * cos(30.0 degrees)
W = (F_applied * √3/2) * 3.0 m * √3/2
Therefore, the work done by the force on the block is (F_applied * √3/2) * 3.0 m * √3/2 Joules.
Part 2:
To find the work done by gravity on the block, use the formula W = m * g * h, where m is the mass of the block, g is the acceleration due to gravity, and h is the height.
In this case, the block is being pushed vertically upwards, so the height h is 3.0 m. The mass of the block is given as 3.0 kg, and the acceleration due to gravity is 9.81 m/s^2.
W = m * g * h
W = 3.0 kg * 9.81 m/s^2 * 3.0 m
Therefore, the work done by gravity on the block is 3.0 kg * 9.81 m/s^2 * 3.0 m Joules.
Part 3:
The magnitude of the normal force between the block and the wall can be found using Newton's second law: ΣF = m * a, where ΣF is the sum of the forces in the vertical direction, m is the mass of the block, and a is the acceleration in the vertical direction.
In this case, the forces acting vertically are the normal force and the gravitational force. Since the block is moving vertically up at a constant velocity, the net force in the vertical direction is zero.
ΣF = Normal force - mg = 0
Therefore, the normal force is equal to the weight mg.
The weight is given by mg = m * g = 3.0 kg * 9.81 m/s^2.
Therefore, the magnitude of the normal force between the block and the wall is 3.0 kg * 9.81 m/s^2 Newtons.
1. Fb = 3kg * 9.8N/kg = 29.4N @ 30deg =
Force of block.
Fp = 29.4sin30 = 14.7N = Force parallel to plane acting downward.
Fv = 29.4cos30 = 8.82N = Force perpendicular to plane acting downward.
Ff = u*Fv = 0.3 * 8.82 = 2.65N. = Force of friction.
Fap - Fp - Ff = 0,
Fap = Fp + Ff = 14.7 + 2.65 = 17.35N.
= Force applied parallel to plane
acting upward.
W = Fap * d = 17.35 * 3 = 52.05J.