How much work is needed to push a 117- kg packing crate a distance of 2.75 m up an inclined plane that makes an angle of 28 o with the horizontal, when inclined plane is frictionless

Explanation:

The minimum total work is the work needed to counteract the work associated with the weight:

Given Information:  

mass of crate = m = 117 kg

distance = d = 2.75 m  

Inclined plane angle = θ = 28°

Required Information:  

Work done = W = ?

Work done = 1480.3 N.m

Explanation:

Refer to the attached image, since there is no friction, the only force acting on the crate is the force in the downward direction given by

Fy = mgsin(θ)

Fy = 117*9.8*sin(28°)

Fy = 538.3 N

Now as we know work done is given by

W = Fy*d

W = 538.3*2.75

W = 538.3*2.75

W = 1480.3 N.m

Therefore, 1480.3 N.m of work is required to push a 117 kg  of crate to a distance of 2.75 m up an inclined plane making an angle of 28° with the horizontal surface.

f is prop to m1xm2 double m1 double force

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