Answer: The correct answer is option B.
Explanation:
Mass of the sled = 10 kg
Initial speed of the sled = 2 m/s
Kinetic energy of the sled = 
Work done by the sled = 20 joules
The work done by the friction will be in opposite direction and equal to the magnitude of the work done of the sled that - 20 J.
Hence, correct answer is option B.
Answer:
I would go with 2
Explanation:
But i would also not go with my answer. Lol
Mohs hardness scale. hope this helps
60 RPM equals one hertz (i.e., one revolution per second, or a period of one second). The SI unit for period is the second.
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>
