A man pushing a crate of mass m = 92.0 kg at a speed of v = 0.845 m/s encounters a rough horizontal surface of length scripted 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 283 N on the crate, find the following.
(c) Find the speed of the crate when it reaches the end of the rough surface.
I already found the net force to be 3N and the work done on the crate while its on the rough surface to be -25.4 J.
To find the final speed I attempted to use work=mass*distance*acceleration to solve for a and I got a=.424
Then I used v(final)^2=v(inital)^2+2a*deltax
my answer for this was incorrect.
Can someone please explain why this was the incorrect approach and how I should have gone about solving the problem? Thanks!
how did you get a net force of 3N?
net force=283-mg*.357 ?
Is this thing going downhill? with a horizontal force?
I am lost on what is happening here.
To find the final speed of the crate when it reaches the end of the rough surface, you can use the work-energy theorem. The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy.
In this case, the work done on the crate consists of two parts: the work done by the applied force and the work done by the friction force.
1. Work done by the applied force:
The work done by the applied force is given by the formula W = F * d * cos(theta), where F is the magnitude of the force, d is the displacement, and theta is the angle between the force and the displacement. In this case, the force applied is 283 N and the displacement is 0.65 m. Since the force and the displacement are in the same direction, theta is 0 degrees. Therefore, the work done by the applied force is W_applied = 283 N * 0.65 m * cos(0) = 183.95 J.
2. Work done by the friction force:
The work done by the friction force is given by the formula W_friction = -u * N * d, where u is the coefficient of kinetic friction, N is the normal force, and d is the displacement. The normal force can be calculated as N = m * g, where m is the mass of the crate and g is the acceleration due to gravity (approximated as 9.8 m/s^2). In this case, the mass is 92.0 kg, the coefficient of kinetic friction is 0.357, and the displacement is 0.65 m. Therefore, the normal force is N = 92.0 kg * 9.8 m/s^2 = 901.6 N. Substituting these values into the formula, we get W_friction = -0.357 * 901.6 N * 0.65 m = -166.35 J.
3. Total work done on the crate:
The total work done on the crate is the sum of the work done by the applied force and the work done by the friction force. Therefore, W_total = W_applied + W_friction = 183.95 J + (-166.35 J) = 17.6 J.
According to the work-energy theorem, this work done on the crate is equal to the change in its kinetic energy. Therefore, we can write the equation as:
W_total = delta(KE) = (1/2) * m * (v_final^2 - v_initial^2),
where v_final is the final velocity (which we want to find), v_initial is the initial velocity (given as 0.845 m/s), and m is the mass of the crate (given as 92.0 kg).
Substituting the known values into the equation, we have:
17.6 J = (1/2) * 92.0 kg * (v_final^2 - (0.845 m/s)^2).
Simplifying the equation:
17.6 J = 46.0 kg * (v_final^2 - 0.713 m^2),
17.6 J / 46.0 kg = v_final^2 - 0.713 m^2,
0.382 J/kg = v_final^2 - 0.713 m^2.
Rearranging the equation to solve for v_final^2:
v_final^2 = 0.382 J/kg + 0.713 m^2,
v_final^2 = 1.095 J/kg,
v_final = sqrt(1.095 J/kg).
Now you can calculate the final velocity by taking the square root of 1.095 J/kg using a calculator:
v_final ≈ 1.046 m/s.
Therefore, the speed of the crate when it reaches the end of the rough surface is approximately 1.046 m/s.