a) Fcos15 d1
b) W = 1/2mv^2
c) 1/2mv^2 = mu(mg) d2
d1 = 12, d2 =15
b) W = 1/2mv^2
c) 1/2mv^2 = mu(mg) d2
d1 = 12, d2 =15
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 (Ï€/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.