Asked by sadie
a sled mass of 50kg is pulled horizontally over flat ground. the static friction coefficient is .30, and the sliding friction coefficient is .10.
what minimum amount of force must be applied to the sled in order to start it moving? i got 147.15N
what amount of applied force will keep it moving at a constant velocity of 3.0m/s?
what minimum amount of force must be applied to the sled in order to start it moving? i got 147.15N
what amount of applied force will keep it moving at a constant velocity of 3.0m/s?
Answers
Answered by
Henry
a. M*g = 50 * 9.8 = 490 N. = Wt. of sled = Normal force(Fn).
Fs = us*Fn = 0.3 * 490 = 147 N.
Fap-Fs = M*a.
Fap = M*0 + 147 = 147 N.
b. Fk = uk*Fn = 0.10 * 490 = 49 N.
Fap = M*0 + 49 = 49N.
Fs = us*Fn = 0.3 * 490 = 147 N.
Fap-Fs = M*a.
Fap = M*0 + 147 = 147 N.
b. Fk = uk*Fn = 0.10 * 490 = 49 N.
Fap = M*0 + 49 = 49N.
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