FuLL SCREEN PRINTrRVERStow -BACK NEXT MESSAGE MY INSTRUCTOR A 77.8-kg person, ru

Question

FuLL SCREEN PRINTrRVERStow -BACK NEXT MESSAGE MY INSTRUCTOR A 77.8-kg person, running horizontalily with a velocity of +3.80 m/s, jumps onto a 15.0-kg sled that is initially at rest. (a) ignoring the effects of friction during the collision, find the velocity of the sled and person as they move away. (b) The sled and person coast 30.0 m on level snow before coming to rest. What is the coefficient of kinetic friction between the sled and the snow? Chapter 07, Problem 44 (a) Number Units (b) Number Units

Explanation / Answer

By conserving momentum, 77.8*3.8= (77.8+15)*v

Or, 295.64= 92.8*v

Or, v= 3.18m/s

So, (a) velocity of the sled and person as they move away is v= 3.18 m/s

Now, using conservation of energy, initial kinetic energy is completely converted to heat loos due to friction. So, 0.5*M*v^2= F*x { where F= u*N= u*Mg= u*92.8*9.8= 909.44 }

0.5*92.8*3.18^2= F*30

Or, F= 15.64

Or, u*909.44= 15.64

So, u= 0.0172

So, (b) the coefficient of kinetic friction between the sled and the snow is 0.0172.

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