A box that weighs 5.00×10^2 N is sliding down a ramp at a constant speed. The angle the ramp makes with the horizontal is 25°. What is the coefficient of friction between the box and the ramp?

0.466 (3 sig. fig.)

Explanation:

Frictional force acting on the box = 5.00×10^2xsin25

Normal force acting on the box = 5.00×10^2xcos25

coefficient of friction = 0.466 (3 sig. fig.)

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft

answergravity;