To give a 19 kg child a ride, two teenagers pull on a 3.7 kg sled with ropes. Bo

Question

To give a 19 kg child a ride, two teenagers pull on a 3.7 kg sled with ropes. Both teenagers pull with a force of 55 N at an angle of 35 relative to the forward direction, which is the direction of motion. In addition, the snow exerts a retarding force on the sled that points opposite to the direction of motion, and has a magnitude of 57 N.

What is the friction coefficient µk of the sled on the snow?

I've already calculated the acceleration of the sled and child to be 1.5 m/s2.

I'm thinking the answer may be 33.11N/57N = .58 but I'm not positive. Any help is greatly appreciated.

Explanation / Answer

Here ,

retarding force , Fr = 57 N

force by teeenager , F = 55 N

let the normal force is N

N = mg - 2 * Fr * sin(35)

N = 19 * 9.8 -2 * 55 * sin(35)

N = 123 N

Noew , for coefficient of friction is uk

uk * N = Fr

57 = uk * 123

uk = 0.468

the friction coefficient µk of the sled on the snow is 0.468

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