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jimmy4
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I have attempted this question, please feel feel to comment.
A horizontal force of 80 N acts on a mass of 6 kg resting on a horizontal surface. The mass is initially at rest and covers a distance of 5m in 0.92 s under the action of the force. Assuming that there is no energy lost to air resistance and therefore that the acceleration is constant.
1. calculate the total energy expended in the acceleration
2. calculate the coefficient of friction between the mass and the surface.
Ans attempt.
1. acceleration s = (1/2) at^2
s = (1/2) a(0.92)^2
5 = 0.4232a
a = 5/0.4232
a = 11.815 m/s
Final Velosity v = a * t.
v = 11.815 * 0.92 = 10.87ms^2
Initial Velosity u = 11.815 * 0 = 0
Kinetic Energy Ke = 1/2 (mv^2)
= (1/2) 6 * 10.87^2
= 354.46 Joules
2. Force = m * a the fifferance between 80 N and the friction.
6 * 11.815 = 70.89 N
P = 80 - 70.79 = 9.11 N
P = (mu)Rn
mu = P/Rn
9.11/58.86
mu = 0.154
Any advice would be welcome. Jimmy
A horizontal force of 80 N acts on a mass of 6 kg resting on a horizontal surface. The mass is initially at rest and covers a distance of 5m in 0.92 s under the action of the force. Assuming that there is no energy lost to air resistance and therefore that the acceleration is constant.
1. calculate the total energy expended in the acceleration
2. calculate the coefficient of friction between the mass and the surface.
Ans attempt.
1. acceleration s = (1/2) at^2
s = (1/2) a(0.92)^2
5 = 0.4232a
a = 5/0.4232
a = 11.815 m/s
Final Velosity v = a * t.
v = 11.815 * 0.92 = 10.87ms^2
Initial Velosity u = 11.815 * 0 = 0
Kinetic Energy Ke = 1/2 (mv^2)
= (1/2) 6 * 10.87^2
= 354.46 Joules
2. Force = m * a the fifferance between 80 N and the friction.
6 * 11.815 = 70.89 N
P = 80 - 70.79 = 9.11 N
P = (mu)Rn
mu = P/Rn
9.11/58.86
mu = 0.154
Any advice would be welcome. Jimmy