Asked by denise

A 10-kg box is pulled along a horizontal surface by a
force of 40N applied at a 30° angle above
horizontal. The coefficient of kinetic friction is 0.3.
Calculate the acceleration
Answered by Henry
Wb = M*g = 10*9.8 = 98 N.

Fp = 98*sin 0 = 0 = Force parallel to the surface.

Fn = 98*Cos 0 - 40*sin30 = 78 N. = Normal force.

Fk = u*Fn = 0.3 * 78 = 23.4 N. = Force of kinetic friction.

a=(Fap*Cos30-Fp-Fk)/M=(34.6-0-23.4)/10 = 1.12 m/s^2.
Answered by jack
20
Answered by ja
Find Acceleration if coefficient of friction is 0.2. and applied force is 39.6 .
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