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oxon88
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Homework Statement
A horizontal force of 80N acts on a mass of 6Kg resting on a horizontal surface. The mass is initially at rest and covers a distance of 5m in 0.92s under the action of the force. Assuming there are no energy losses due to air resistance and therefore that the acceleration is constant
i) calculate the total energy expended in the acceleration
ii) calculate the coefficient of friction between the mass and the surface.
Homework Equations
v = s/t
a = (v-u) / t
s = ut + (1/2)a.t2
F = m.a
Ke = 1/2*m*V^2
The Attempt at a Solution
first I calculated the acceleration:s = (1/2).a.t2
5 = (1/2)a.0.922
10/0.922 = a = 11.815 ms-2Then i calculated the net force:
F = m.a
F = 6 * 11.815 = 70.89N
therefore i think friction = 80N - 70.89N = 9.11N Would this be the coefficient of friction?Next i calculated the speed after 0.92s. s = d/t = 5/0.92 = 5.435 ms-1
Then calculate kinetic energy: = (1/2)m.v2 = 3*5.4352 = 88.62Jso now would the total energy expended = kinetic energy + Friction force * 5m?
= 88.62 + 9.11 * 5 = 488.65 Nm
can anyone verify this? and let me know if I'm way off please?