Pete slides a crate up a ramp with constant speed at an angle of 24.3 ◦ by exerting a 289 n force parallel to the ramp. how much work has been done against gravity when the crate is raised a vertical distance of 2.39 m? the coefficient of friction is 0.33. answer in units of j.

W=972.83 J

Explanation:

Given,

angle of inclination of ramp = θ = 24.3°

Force exerted on the block = F= 289 N

Crate is raised to height of = 2.39 m

coefficient of friction = 0.33

where g is the acceleration due to gravity = g = 9.8 m/s²

Work done = ?

let m be the mass of the crate

Gravity force along ramp = m g sinθ

Friction force =  μm g cosθ

now writing all the force

F = m g sinθ  + μm g cosθ

By putting the values

F = m g (sinθ  + μcosθ)              

289 = 9.8 x m  ( sin 24.3°+ 0.33 cos 24.3°)                 ( take g =10 m/s²)

289 = 9.8 m(0.71)

m = 41.53 kg

Work done will be equal to

W= m g h

W= 41.53 x 9.8 x 2.39 J

W=972.83 J

i believe the answer is a

c)as speed changes, wavelength and frequency change.

everything changes!