Explanation:
The angle of the handle relative to the horizontal is 35°. The angle of the ramp to the horizontal is 7°. So the angle of the handle relative to the ramp is 28°.
cos 28° = 50 / F
F = 50 / cos 28°
F = 56.6 lbs
Answer:
n = 5 approx
Explanation:
If v be the velocity before the contact with the ground and v₁ be the velocity of bouncing back
= e ( coefficient of restitution ) = 
and
h₁ is height up-to which the ball bounces back after first bounce.
From the two equations we can write that
So on
= .00396
Taking log on both sides
- n / 2 = log .00396
n / 2 = 2.4
n = 5 approx
Answer:
See below
Explanation:
Normal force = m g cos 53 = 8 kg * 9.8 m/s^2 * cos 53 = 47.1823 N
no work is done by this force
Force friction = coeff friction * force normal = .4 * 47.1823 = 7.55 N
work of friction = 7.55 * 2 m = 15.1 j
Force Downplane = mg sin 53 = 62.61 N
work = 62.61 * 2 = 125.22 j
Net Force downplane = force downplane - force friction = 55.06 N
net Work = force * distance = 55.06 N * 2 M = 110.12 j
