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
(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)
To solve the given problem, we will break it down into different steps:
Step 1: Determine the forces acting on the steel block.
- The force of gravity acting vertically downward can be determined by multiplying the weight of the block by the acceleration due to gravity (9.8 m/s^2). In this case, the weight of the block is 12 N, so the force of gravity is 12 N * 9.8 m/s^2 = 117.6 N.
- The force pulling the block up the incline can be determined by resolving the given force into components. The vertical component is given by 7.35 N * sin(10°) = 1.28 N, and the horizontal component is given by 7.35 N * cos(10°) = 7.16 N.
- The force of friction can be determined by multiplying the coefficient of friction (0.20) by the normal force. The normal force is equal to the component of the weight perpendicular to the inclined plane, which can be calculated as 12 N * cos(20°) = 10.91 N. Thus, the force of friction is 0.20 * 10.91 N = 2.18 N.
Step 2: Calculate the work done on the steel block.
- The work done on an object can be determined by multiplying the force applied on the object by the displacement. In this case, the force applied is the horizontal component of the pulling force (7.16 N) and the displacement is 2.0 m. Therefore, the work done is 7.16 N * 2.0 m = 14.32 J.
Step 3: Calculate the increase in potential and kinetic energy.
- The increase in potential energy can be determined by calculating the change in height. The change in height (Δh) can be calculated as the vertical displacement (2.0 m) multiplied by the sine of the angle of the incline (20°). Thus, Δh = 2.0 m * sin(20°) = 0.684 m. The increase in potential energy is then given by ΔPE = m * g * Δh = 12 N * 9.8 m/s^2 * 0.684 m = 80.55 J.
- The increase in kinetic energy can be calculated by knowing that the initial velocity is 0 m/s (as the block starts from rest) and the final velocity can be calculated using the work-energy principle: ΔKE = W - ΔPE - Wfriction. Therefore, ΔKE = 14.32 J - 80.55 J - (2.18 N * 2.0 m) = -69.41 J.
Step 4: Calculate the amount of work required to overcome the friction force.
- The work required to overcome the friction force is given by Wfriction = force of friction * displacement = 2.18 N * 2.0 m = 4.36 J.
Therefore, the solutions to the given problem are:
(a) The work done on the steel block is 14.32 J.
(b) The increase in potential energy is 80.55 J and the increase in kinetic energy is -69.41 J.
(c) The amount of work required to overcome the friction force is 4.36 J.