A 2 kg block is pushed 2 m along a fixed horizontal surface by a horizontal force of 20 N. The coefficient of kinetic friction of the surface is µk= 0.40.
(a) How much work is done by the 20 N force?
(b) what is the work done by the friction force?
(c) What is the total work done by the net force on the object?
Wb = mg = 2kg * 9.8N/kg = 19.6N. = Weight of block.
Fb = 19.6N. @ 0deg. = Force of block.
Fp = 19.6sin(0) = 0 = Force parallel to
surface.
Fv = 19.6cos(0) = 19.6N. = Force perpendicular o surface.
Ff = u*Fv = 0.4 * 19.6N. = 7.84N. = Force of friction.
a. W = Fap * d = 20 * 2 = 40J.
b. W = Ff * d = 7.84 * 2 = 15.68J.
c. Fn = Fap - Fp - Ff,
Fn = 20 - 0 - 7.84 = 12.16N.
W = Fn * d =12.16 * 2 = 24.32J.
To answer these questions, we need to understand the concept of work.
Work is defined as the product of the force applied on an object and the distance over which the force is applied. Mathematically, work is given by the equation W = F * d * cos(θ), where W is work, F is force, d is distance, and θ is the angle between the force and the direction of motion.
Now let's address each question:
(a) How much work is done by the 20 N force?
The force applied by the 20 N force is in the same direction as the displacement of the block. Therefore, the angle between the force and the direction of motion is 0 degrees (θ = 0). In this case, the work done by the force can be calculated using the equation W = F * d * cos(θ). Plugging in the values, we get W = 20 N * 2 m * cos(0°), which simplifies to W = 40 J. Therefore, the work done by the 20 N force is 40 Joules.
(b) What is the work done by the friction force?
The work done by friction can be calculated using the equation W = F * d * cos(θ). However, in this case, friction acts in the opposite direction to the motion, which means the angle between the force and the direction of motion is 180 degrees (θ = 180°). Therefore, the work done by friction can be calculated as W = F * d * cos(180°). Since the cosine of 180° is -1, the equation simplifies to W = -F * d. Plugging in the values, we get W = -(µk * Mg) * d, where µk is the coefficient of kinetic friction, M is the mass of the block (2 kg), g is the acceleration due to gravity (approximately 9.8 m/s^2), and d is the distance moved (2 m). Thus, W = -(0.40 * 2 kg * 9.8 m/s^2) * 2 m, which simplifies to W = -15.68 J. Therefore, the work done by the friction force is -15.68 Joules.
(c) What is the total work done by the net force on the object?
The total work done by the net force on the object is the algebraic sum of the work done by the 20 N force and the work done by the friction force. Since work done against friction is considered negative, we can sum up the work done by the 20 N force (40 J) and the work done by the friction force (-15.68 J) to calculate the total work. In this case, the total work is 40 J + (-15.68 J), which simplifies to 24.32 J. Therefore, the total work done by the net force on the object is 24.32 Joules.