work=F.distance=force*distance*cosAngle between them
= mg*30*cos(90-28)
A 1.11 kg block slides down a 30.0 m long 28.0° incline at constant velocity. How much work is done by friction?
2 answers
M*g = 1.11 * 9.8 = 10.9 N. = Wt. of block.
Fp = 10.9*sin 28 = 5.1 N. = Force parallel to incline.
Fn = 10.9*cos28 = 9.6 N. = Normal force.
u*Fn = Force of kinetic friction
Fp - u*Fn = M*a. a = 0(constant velocity).
5.1 - u*Fn = 1.11*0,
u*Fn = 5.1 N. = force of kinetic friction.
Work = 5.1 * 30 = 153 Joules.
Fp = 10.9*sin 28 = 5.1 N. = Force parallel to incline.
Fn = 10.9*cos28 = 9.6 N. = Normal force.
u*Fn = Force of kinetic friction
Fp - u*Fn = M*a. a = 0(constant velocity).
5.1 - u*Fn = 1.11*0,
u*Fn = 5.1 N. = force of kinetic friction.
Work = 5.1 * 30 = 153 Joules.