A 15 kg crate is pulled at constant speed a distance d= 5.7m up a frictionless ramp, to a

height h of 2.5 m above its starting point.
a) What force F must the cable exert on the
crate?
b) How much work is done on the crate by the
force F?

M*g = 15 * 9.8 = 147N. = Wt. of crate.

sin A = h/d = 2.5m/5.7m = 0.43860
A = 26o.

a. F = Fp = 147*sin A = Force parallel
with the incline.

b. Work = F*d =
d = 5.7m.
Work in Joules.

To find the force F exerted by the cable on the crate, we need to resolve two components of forces acting on the crate: the gravitational force and the force component parallel to the ramp.

a) The gravitational force acting on the crate is given by the formula F_gravity = m * g, where m is the mass of the crate and g is the acceleration due to gravity (approximately 9.8 m/s²). Plugging in the values, we have:

F_gravity = 15 kg * 9.8 m/s² = 147 N

The force component parallel to the ramp is equal to the force of gravity projected onto the ramp. Since the crate is pulled up the ramp at a constant speed, the force component parallel to the ramp must be equal in magnitude and opposite in direction to the force of gravity. Therefore, the force exerted by the cable on the crate is also 147 N.

b) The work done on the crate by the force F can be calculated using the formula: Work = force * distance * cos(theta), where theta is the angle between the force and the direction of motion.

In this case, the force F and the displacement d are parallel, so cos(theta) = 1. Therefore, the work done on the crate is given by:

Work = F * d = 147 N * 5.7 m = 838.9 J

So, the work done on the crate by the force F is 838.9 Joules.

To find the answers, we need to use the concepts of work and energy.

a) First, let's find the gravitational potential energy gained by the crate. The formula for gravitational potential energy is:

PE = m * g * h

where PE is the potential energy, m is the mass of the crate, g is the acceleration due to gravity, and h is the height gained. In this case, since the crate is being pulled up a ramp, we take h as the vertical distance gained.

PE = 15 kg * 9.8 m/s^2 * 2.5 m
PE = 367.5 J

Since the crate is being pulled at a constant speed, the work done by the force of the cable must be equal to the potential energy gained by the crate. The formula for work is:

Work = F * d

where F is the force applied, and d is the distance over which the force is applied.

Therefore, to find the force F, we rearrange the equation:

F = Work / d

Substituting the given values:

F = 367.5 J / 5.7 m
F ≈ 64.47 N

Therefore, the force F that the cable must exert on the crate is approximately 64.47 N.

b) The work done on the crate by the force F is given by the formula:

Work = F * d

Substituting the given values:

Work = 64.47 N * 5.7 m
Work ≈ 367.5 J

Therefore, the work done on the crate by the force F is approximately 367.5 J.