(a) The work done on the steel block is given by the product of the force applied and the displacement of the block in the direction of the force. Since the force is applied at an angle with the direction of displacement, we need to find the component of the force in the direction of displacement.
Component of force in the direction of displacement = force x cos(angle between force and displacement)
= 7.35 N x cos(10°)
= 7.17 N (approx)
Therefore, work done on the steel block = force x displacement x cos(angle between force and displacement)
= 7.17 N x 2.0 m x cos(0°) [since the force and displacement are in the same direction]
= 14.34 J
(b) The increase in potential energy is given by the product of the weight of the block and the vertical height it is lifted.
Increase in potential energy = weight x height
= 12N x 2.0 m x sin(20°) [since the height lifted is along the vertical direction, which is perpendicular to the inclined plane]
= 4.12 J (approx)
The final kinetic energy of the block can be found using the work-energy principle, which states that the net work done on an object is equal to its change in kinetic energy.
Net work done = work done on the steel block - work done by friction
= 14.34 J - work done by friction
Since the block is starting from rest, its initial kinetic energy is zero. Therefore, the total change in kinetic energy is equal to the final kinetic energy.
Change in kinetic energy = final kinetic energy - initial kinetic energy
= final kinetic energy - 0
= final kinetic energy
Therefore, final kinetic energy = net work done = 14.34 J - work done by friction.
(c) The work required to overcome the friction force is given by the product of the friction force and the displacement of the block in the direction opposite to the force.
Work done by friction = friction force x displacement x cos(180°)
= 0.20 x 12N x 2.0m x cos(180° - 20°) [since the displacement is in the direction opposite to the friction force and the angle between them is 180° - 20°]
= 4.72 J (approx)
Therefore, final kinetic energy = net work done - work done by friction
= 14.34 J - 4.72 J
= 9.62 J (approx)
A steel block weighing 12N is pulled up an incline plane 20° above the horizontal by a constant force of 7.35 N which makes an angle of 10° above the inclined plane. The block stars from rest and is pulled 2.0m along the inclined plane. the coefficient of friction between the block and inclined plane is 0.20
(a) find the work done on the steal
(b) find the increase in potential and Kinetic Energy
(c) what Is the amount of work required to overcome the friction force
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