A 5kg box is pulled for a distance of 12m with a force of 20N along frictionless surface. The force is applied at an angle that is 15 deg. above the horizontal. a.) how much work was done by the applied force? b) how fast will the box be moving at the end of the 12m if it started from rest? c.) if box experiences frictional surfact at end of first 12m and 20N force is no longer acting on object, what must coeff. of kin.friction in order for box to come to rest in exactly 15m.
a) Fcos15 d1
b) W = 1/2mv^2
c) 1/2mv^2 = mu(mg) d2
d1 = 12, d2 =15
To find the answers to these questions, we need to use the concepts of work, energy, and friction.
a) Work done by the applied force:
Work = Force x Distance x cos(angle)
Given: Force = 20N, Distance = 12m, Angle = 15 degrees
Convert the angle to radians: 15 degrees x (π/180) = 0.2618 radians
Work = 20N x 12m x cos(0.2618 radians) ≈ 233.3 Joules
b) Velocity of the box at the end of 12m:
To find the final velocity, we can use the work-energy principle.
The work done on the box is equal to the change in its kinetic energy.
Work = ΔKE (change in kinetic energy)
Since the box started from rest, the initial kinetic energy is 0.
Work = KE final - KE initial
Work = 0.5mv^2 - 0 (initial KE is 0)
Work = 0.5 x 5kg x v^2 (mass of the box is 5kg)
From part a, we found that the work done is approximately 233.3 Joules.
233.3 = 0.5 x 5kg x v^2
v^2 = 233.3 J / (0.5 x 5kg) ≈ 46.7 m^2/s^2
v ≈ √46.7 ≈ 6.8 m/s (approximate)
c) Coefficient of kinetic friction:
To determine the coefficient of kinetic friction, we need to use the work-energy principle again.
The work done by friction will be equal to the work done by the applied force to bring the box to rest.
Work of friction = Force of friction x Distance
Given: Force of friction = 20N (opposite to the applied force), Distance = 15m
We know that the work done by the applied force is 233.3 J (from part a).
So, Work of friction = 233.3 J
The work done by friction can also be expressed as:
Work of friction = Force of friction x Distance x cos(180 degrees)
Work of friction = Force of friction x Distance x (-1) (cos(180 degrees) is -1)
Therefore, -20N x 15m = 233.3 J
-300N x m = 233.3 J
Dividing both sides by -300N:
m = 233.3 J / (-300N) ≈ -0.78 m
The negative sign indicates the direction opposite to the applied force. However, the coefficient of friction cannot be negative, so we take the absolute value:
|μ| = |-0.78| ≈ 0.78
Therefore, the coefficient of kinetic friction must be approximately 0.78 for the box to come to rest in exactly 15m.