A 13.2 kg crate is pulled by a 157 N force (parallel to the incline) up a rough incline. Assume that the incline makes an angle of 34.0° with the horizontal and that the coefficient of kinetic friction is 0.434. The crate has an initial speed of 1.53 m/s. It is pulled a distance of 6.30 m. What is the change in gravitational potential energy of the crate? Answer in units of J.
I thought that the work done by the force of gravity would be equal to the change in gravitational potential energy, (-465), but it turned out to be wrong. Any suggestions?
6.3 * sin 34 = 3.52 meters up
m g h = 13.2 * 9.81 * 3.52 = 456
so I get
-456 J
I think you reversed digits
To calculate the change in gravitational potential energy of the crate, we need to consider the work done against gravity. The work done by the force of gravity is given by the formula W = mgh, where m is the mass, g is the acceleration due to gravity, and h is the change in height.
In this case, the crate is not moving vertically, so there is no change in height. Therefore, the work done by the force of gravity is zero, and there is no change in gravitational potential energy.
Instead, we need to calculate the work done against friction. The work done against friction can be found using the formula W = Fd, where F is the force of friction and d is the distance over which the force is applied.
First, let's calculate the force of friction. The force of friction can be calculated using the formula F_friction = μ_k * F_n, where μ_k is the coefficient of kinetic friction and F_n is the normal force.
The normal force can be found using the formula F_n = mg * cos(θ), where θ is the angle of the incline.
F_n = (13.2 kg) * (9.8 m/s^2) * cos(34°)
F_n = 130.52 N
Now, let's calculate the force of friction:
F_friction = (0.434) * (130.52 N)
F_friction = 56.57 N
Next, let's calculate the work done against friction:
W = (56.57 N) * (6.30 m)
W ≈ 356.39 J
Therefore, the change in gravitational potential energy of the crate is approximately 356.39 J.