1. A person pushes 20-kg box 2.0m up a 16º incline at a constant speed. If his f
ID: 2007034 • Letter: 1
Question
1.A person pushes 20-kg box 2.0m up a 16º incline at a constant speed. If his force is parallel to the incline and the coefficient of kinetic friction is 0.3, calculate (a) the minimum force this man needs, (b) the work done by the man, (c) the work done by the force of gravity, (d) the work done by the frictional force, (e) the work done by the normal force, and (f) the net work.
2.
A 1200-kg car uses a power of 110 hp to climb an inclined surface at a constant speed of 65 km/h. If the frictional forces add up to be 900 N, determine the angle of the incline above the horizontal.
3.
At the bottom of a ramp, a 15-kg crate moves at a speed of 5 m/s and slides 1.8 m up the ramp, stops, and slides down. If the ramp has an angle of 30º above the horizontal, determine (a) the frictional force, (b) the coefficient of kinetic friction, and (c) the speed when the crate reaches the bottom.
4.
A 2.0-kg rock is thrown vertically with a speed of 15 m/s. Determine (a) the maximum height the rock can reach, (b) the potential energy when the rock reaches one third of the maximum height from the ground, and (c) the speed at this height. Use the energy process for this problem.
5.
A spring is vertically placed on a flat surface. An object of 5 kg held at a height of 0.5 m from the top of the spring falls directly onto the spring. If the spring constant k = 1200 N/m, determine the magnitude the spring can be compressed by the object.
6.
An object is acted by a varying force. The force linearly increases from zero to 12 N in the first 6 meters, and linearly decreases from 12 N to zero in the last 4 meters. (a) To establish the equations of the force for these two cases and (b) to determine the net work of these two cases.
Explanation / Answer
(a) Applying the Newtom's 2nd law of motion, F = mg(sin16+cas16) = 20*9.8(0.275+0.288) = 110.54 N the minimum force is 110.54 N. (b) the displacement is d = 2/sin16 = 7.255 m the wotk wone by the man, w = F*d = 110.54*7.255 w = 802.06 J (c) the work done by the gravity, Wg = mgh = 20*9.8*2 Wg = 392 J (d) the work done by the frictional force , Wf = mgcas16 = 0.3*20*9.8*cas16 = 56.52 J (e) the work done by the noemal force, Wn = 0 J since force is perpendicular to displacement. (f) the net work done, W' = work done by the man- work done by the frictional force = 802.06-56.52 = 745.53 J * Plz post each question separately.Related Questions
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