Answer:
Hence, work done= 287.54 J
Explanation:
Given data:
angle of ramp with the ground θ =20°
force applied = 76 N
work done on the crate to slide down 4 m down the ramp
W= F×d cosθ ( only the cos component of the force will slide the crate down)
W= 76×4×cos20= 287.54 J
Here in this question as we can see there is no air friction so we can use the principle of energy conservation


now here we know that



now plug in all values in above equation

divide whole equation by mass "m"



so height of the ball from ground will be 1.35 m
Given
Weight of the block A, Wa = 20 lb, weight of block B Wb = 50 lb. Applied
force to block A, P = 6lb, coefficient of static friction µs = 0.4, coefficient
of kinetic friction µk = 0.3. If a force P
is applied to the body, no relative motion will take place until the applied
force is equal to the force of friction Ff, which is acting opposite to the
direction of motion. Magnitude of static force of friction between block A and
block B, Fs = µsN, where N is
reaction force acting on block A. Now, resolve the forces Fx = max. P = (mA +
mB)a,
6 = (20 / 32.2 + 50 / 32.2)a
2.173a = 6
A = 2.76 ft/s^2
To check slipping occurs between block A and block B, consider block A:
P – Ff = mAaA
6 – Ff = 1.71
Ff = 4.29 lb
And also,
N = wA. We know static friction,
Fs = µsN
Fs = 0.4 x 20
Fs = 8lb
Frictional force is less than static friction. Ff < Fs
<span>Therefors, acceleration of block A, aA = 2.76 ft/s^2, acceleration of
block B aB = 2.76 ft/s^2</span>
Answer:
filament bulbs give off lots of metal filament that transfers wasted energy to the surrounding.