Answer:
(a) 81.54 N
(b) 570.75 J
(c) - 570.75 J
(d) 0 J, 0 J
(e) 0 J
Explanation:
mass of crate, m = 32 kg
distance, s = 7 m
coefficient of friction = 0.26
(a) As it is moving with constant velocity so the force applied is equal to the friction force.
F = 0.26 x m x g = 0.26 x 32 x 9.8 = 81.54 N
(b) The work done on the crate
W = F x s = 81.54 x 7 = 570.75 J
(c) Work done by the friction
W' = - W = - 570.75 J
(d) Work done by the normal force
W'' = m g cos 90 = 0 J
Work done by the gravity
Wg = m g cos 90 = 0 J
(e) The total work done is
Wnet = W + W' + W'' + Wg = 570.75 - 570.75 + 0 = 0 J
The gravitational field is the Force divided by the mass
Call g the gravitational fiel, F the force exerted by the earth and m the mass of the telescope.
g = F / m
g=9.1x10^4 N / 1.1 x 10^4 kg = 8.27 N/kg
Note that the unit N/kg is equivalent to m/s^2
Answer:
1. energy lost in the lever due to friction
3. visual estimation of height of the beanbag
5. position of the fulcrum for the lever affecting transfer of energy
Explanation:
Edge 2021
