A 50kg trunk is pushed 6.0 m at a constant speed up a 30degree

incline by a constant horizontal force. Thecoefficient of kinetic

energy n/between the trunk and the inclineis 0.20. Calculate the

work done by:

a.) the applied force and

b.) by the weight of the trunk.

c.) How much energy was dissipated by the frictional

forceacting on the trunk?

a) 2034 J

b) - 1471 J

c) - 509 J

Explanation:

The trunk has a mass of 50 kg, so its weight is

f = m * a

f = 50 * 9.81 = 490 N

If the incline is of 30 degrees, the force tangential to the incline is:

ft = f * sin (a)

ft = 490 * sin (30) = 245 N

And the normal force is:

fn = f * cos (a)

fn = 490 * cos (30) = 424 N

The frictional force is:

ff = μ * fn

ff = 0.2 * 424 = 84.8 N

To push the trunk up one must apply a force slightly greater than the opposing forces, the opposing are the tangential component of the weight and the friction force

fp = fn + ff

fp = 245 + 84.8 = 339 N

The work of the applied force is:

L = f * d

Lp = fp * d

Lp = 339 * 6 = 2034 J

The work of the weight is done by the tangential component:

Lw = - 245 * 6 = - 1471 J (it is negative because the weight was opposed to the direction of movement)

The work of the friction force is

Lf = - 84.8 * 6 = - 509 J