Answer:
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
