A50kg 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

unbalanced force

explanation:

neither accurate nor precise.