Some workers do 15000 J of work on a 3500 N crate to push it up a ramp. The ramp is 4 m high. What is the efficiency of the ramp?

93.3%

Explanation:

Efficiency of ramp is given by

Where n is the efficiency, W is work and subscripts o and i represent output and input respectively.

The work input is what the workers do and that is given as 15 kJ

The work output

Since work is product of force and perpendicular distance and substituting force with 3500 N and distance with 4 m we get that the work output is 3500*4=14000 J= 14 kJ

Finally, substituting 14 kJ and 15 kJ for work output and input respectively we deduce that

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