A horizontal force of 150 N is used to push a 33.0 kg packing crate a distance of 6.65 m on a rough horizontal surface. If the crate moves at constant speed and the work done is 997.5 joules. What is the coefficient of kinetic between the crate and the surface.
workin-friction=workdone
150*6.65- mg*mu*6.65=997.5
solve for mu.
To find the coefficient of kinetic friction between the crate and the surface, we need to first understand the concept of work and the forces involved.
1. Work done (W) is given by the formula W = F * d * cos(theta), where F is the applied force, d is the displacement, and theta is the angle between the force and the displacement.
2. In this case, the work done is given as 997.5 joules.
3. Since the crate is moving at a constant speed, the net force acting on it must be zero. The applied force (150 N) is balanced by the force of friction.
4. The horizontal force of 150 N is the force of friction. Therefore, the formula for work done simplifies to W = F * d.
5. Plugging in the values, we have 997.5 joules = 150 N * 6.65 m.
6. Solving for the coefficient of kinetic friction (μ_k), we can use the equation F = μ_k * N, where N is the normal force.
7. Since the crate is on a rough horizontal surface, the normal force (N) is equal to the weight of the crate, which is given by the formula N = m * g, where m is the mass of the crate and g is the acceleration due to gravity (approximately 9.8 m/s^2).
8. Rearranging the equation F = μ_k * N, we have μ_k = F / N.
9. Plugging in the values, we have μ_k = 150 N / (33.0 kg * 9.8 m/s^2).
10. Evaluating the expression, we find that the coefficient of kinetic friction between the crate and the surface is approximately 0.449.