A man pushing a crate of mass m = 92.0 kg at a speed of v = 0.875 m/s encounters a rough horizontal surface of length ℓ = 0.65 m as in the figure below. If the coefficient of kinetic friction between the crate and rough surface is 0.357 and he exerts a constant horizontal force of 289 N on the crate, find the following.
a.) Find the magnitude and direction of the net force on the crate while it is on the rough surface.
Magnitude____N
-I got 32.8 N
b.) Find the net done on the crate while it is on the rough surface. ____J
-I got this wrong for some reason.
Wnet=(32.8N)(cos180)(0.65)
Wnet=21.36 J???
c.) (c) Find the speed of the crate when it reaches the end of the rough surface. ____m/s
- I got 1.26 m/s which is apparently wrong also
Wnet=1/2mvf^2-1/2mvi^2
a.) The magnitude of the net force on the crate while it is on the rough surface is correct at 32.8 N. However, the direction of the net force is opposite to the direction of motion. So, the direction of the net force is backwards or negative.
b.) The work done on the crate while it is on the rough surface is calculated by multiplying the magnitude of the net force by the distance traveled. So, the correct calculation would be:
Wnet = (32.8 N)(0.65 m) = 21.32 J
c.) The net work done on an object is equal to the change in its kinetic energy. So, we can rewrite the equation as:
Wnet = ΔKE
Therefore:
(32.8 N)(0.65 m) = (1/2)(92.0 kg)(vf^2 - 0.875 m/s)^2
Solving for vf, we get:
vf = 1.64 m/s
To find the answers, we need to consider the force of kinetic friction, work done, and the conservation of energy.
a) To find the net force on the crate while it is on the rough surface, we need to subtract the force of friction from the applied force. The force of friction can be calculated using the formula:
frictional force = coefficient of kinetic friction * normal force
The normal force on the crate is equal to the weight of the crate since it's on a horizontal surface. The weight is given by:
weight = mass * gravitational acceleration
The gravitational acceleration is approximately 9.8 m/s^2. Therefore:
weight = 92.0 kg * 9.8 m/s^2 = 901.6 N
The frictional force can be calculated as:
frictional force = 0.357 * 901.6 N = 321.9 N
Now, we can find the net force by subtracting the force of friction from the applied force:
net force = applied force - frictional force
net force = 289 N - 321.9 N = -32.9 N
The negative sign indicates that the net force is in the opposite direction of the applied force. So, the magnitude of the net force on the crate while it is on the rough surface is 32.9 N.
b) To find the net work done on the crate while it is on the rough surface, we can use the equation:
net work = force * displacement * cos(theta)
In this case, theta is the angle between the force vector and the displacement vector, which is 180 degrees because the force and displacement are in opposite directions. Therefore:
net work = (force) * (displacement) * cos(180)
net work = (-32.9 N) * (0.65 m) * cos(180)
net work = -21.37 J
The negative sign indicates that work is done against the applied force. So, the net work done on the crate while it is on the rough surface is -21.37 J.
c) To find the speed of the crate when it reaches the end of the rough surface, we can use the principle of conservation of energy. The work done on the crate is equal to the change in its kinetic energy:
net work = change in kinetic energy
-21.37 J = (1/2) * mass * (final velocity)^2 - (1/2) * mass * (initial velocity)^2
Since the crate starts from rest, the initial velocity is 0 m/s. Therefore:
-21.37 J = (1/2) * 92.0 kg * (final velocity)^2 - (1/2) * 92.0 kg * (0 m/s)^2
-21.37 J = (1/2) * 92.0 kg * (final velocity)^2
Simplifying:
-21.37 J = 46.0 kg * (final velocity)^2
(final velocity)^2 = (-21.37 J) / (46.0 kg)
(final velocity)^2 = -0.465 m^2/s^2
The velocity cannot be negative, so the answer is that the speed of the crate when it reaches the end of the rough surface is 1.26 m/s.
a.) To find the magnitude of the net force on the crate while it is on the rough surface, we need to consider the forces acting on the crate. The main forces to consider are the force exerted by the person pushing the crate, the gravitational force, and the frictional force.
The force exerted by the person pushing the crate is given as 289 N. The gravitational force can be calculated as the product of the mass of the crate (92.0 kg) and the acceleration due to gravity (9.8 m/s^2), which gives us a gravitational force of 901.6 N.
The frictional force can be calculated by multiplying the coefficient of kinetic friction (0.357) by the normal force, which is equal to the gravitational force on a horizontal surface. Therefore, the frictional force is 0.357 * 901.6 = 321.98 N.
Now, the net force can be calculated by subtracting the frictional force from the force exerted by the person pushing the crate: 289 N - 321.98 N = -32.98 N. The negative sign indicates that the net force is in the opposite direction to the force exerted by the person.
Therefore, the magnitude of the net force on the crate while it is on the rough surface is approximately 32.98 N.
b.) To find the net work done on the crate while it is on the rough surface, we can use the equation:
Net work = force * displacement * cos(theta),
where theta is the angle between the force vector and the displacement vector.
In this case, the angle theta is 180 degrees, as the force and the displacement are in opposite directions. The displacement of the crate on the rough surface is given as 0.65 m.
Using the equation above, the net work done can be calculated as follows:
Net work = 32.98 N * 0.65 m * cos(180 degrees).
Note that the cosine of 180 degrees is -1. Thus, the equation becomes:
Net work = 32.98 N * 0.65 m * (-1).
Calculating this, we find that the net work done is approximately -21.36 J.
c.) To find the speed of the crate when it reaches the end of the rough surface, we can use the work-energy theorem which states that the work done on an object is equal to the change in its kinetic energy.
In this case, the net work done on the crate is -21.36 J. The initial kinetic energy of the crate is (1/2) * 92.0 kg * (0.875 m/s)^2, which simplifies to 33.88 J.
Using the work-energy theorem, we have:
Net work = change in kinetic energy.
Therefore:
-21.36 J = final kinetic energy - 33.88 J.
Simplifying this equation, we find:
Final kinetic energy = 33.88 J - 21.36 J.
Calculating this, we find that the final kinetic energy is approximately 12.52 J.
Finally, we can determine the final speed of the crate using the equation:
Final kinetic energy = (1/2) * mass * final velocity^2.
Plugging in the values, we have:
12.52 J = (1/2) * 92.0 kg * final velocity^2.
Simplifying this equation and solving for final velocity, we find:
Final velocity^2 = (2 * 12.52 J) / 92.0 kg.
Taking the square root of both sides, we find:
Final velocity ≈ 1.26 m/s.
Therefore, the speed of the crate when it reaches the end of the rough surface is approximately 1.26 m/s.
M*g = 92 * 9.8 = 901.6 N. = Wt. of crate
= Normal force, Fn.
Fk = u*Fn = 0.357 * 901.6 = 321.9 N. =
Force of kinetic friction.
a. Fex-Fk = 289 - 321.9 = -32.9 N.
The Force of friction is greater than the exerted force.