Now consider friction. The track has a rough surface, the cart has a box tied to

Question

Now consider friction. The track has a rough surface, the cart has a box tied to it and rolls without slipping along this rough surface. Assume that the coefficient of kinetic friction between the box and the plane is Muk = 0.10. The cart is designed to have very little (ie. negligible friction). The inclined plane is set up as shown below. The mass of the cart is 1.09 kg, the mass of the box is 0.540 kg, and the hanging mass is m = 493 g, the angle of the plane is given by 8.9 degree and g = 9.8 ms^-2 . What is the acceleration of the cart along the slope?

Explanation / Answer

Force up the incline = weight of hanging mass = 0.493kg x 9.8 = 4.8314 N
Force down incline = weight component = mg.sin = (Total Weight)*9.8*sin8.9 = (1.09+0.54)kg x 9.8 x sin8.9 = 1.63*9.8*sin8.9 = 15.974 * 0.15471 = 2.4713

Resultant force along incline Fr(before Friction) = (4.8314 - 2.4713) = 2.3601 N (upwards)

Now Friction between the box and the plane = u*m*g*cos8.9 = 0.10*0.54*9.8*cos8.9 = 0.5292*cos8.9 = 0.5228

Resultant force along incline Fr(after Friction) = 2.3601 - 0.5228 = 1.8373 N.

Acceleration = Fr/m = 1.8373N / 1.63kg .. So, a = 1.1271 m/s²

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