A 4.0kg toboggan rests on a frictionless icy surface, and a 2.0kg block rests on

Question

A 4.0kg toboggan rests on a frictionless icy surface, and a 2.0kg block rests on top of the toboggan. The coefficient of static friction between the block and the surface of the toboggan is 0.58, whereas the kinetic friction coefficient is 0.48. The block is pulled by a horizontal force of 30N. a)calculate the block's acceleration b) calculate the toboggan's acceleration c) if the applied force is gradually reduced, at what value do the block and the toboggan have the same acceleration? Please show detailed solutions :)

Explanation / Answer

We expect the motion to be horizontal, so we choss the x axis horizotal and the y axis vertical. The push of 21N has components Fx=Fcos?=(21N)cos35°=17.20N Fy=Fsin?=(21N)sin35°=12.05N Fn=mg+Fy=(4.1kg)(9.8m/s^2)+12.05N=52.23N (be careful when calculating the normal force) Now we apply Newton's second law for the horizontal direction, and include the friction force Fx-Ffr=ma Ffr=µkFn=0.20(52.23N)=10.45N Hence the box does accelerate a=(Fx-Ffr)/m=(17.2N-10.45N)/4.1kg=1.6m/s…

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