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:
The magnitude of the acceleration of the box is 2 m/s².
Explanation:
Given:
Mass of the box, 
kg
Force acting towards east, 
N
Frictional force acting towards west, 
N
Let the acceleration be 
m/s².
Now, net force acting on the box towards east is given as:
From Newton's second law of motion,
Therefore, the magnitude of the acceleration of the box is 2 m/s².
Answer:
3000 kg.m/s
Explanation:
Momentum, p is a product of mass and velocity hence
p=mv where m is mass and v is velocity.
Change in momentum is given by 
where subscripts f and i represent final and initial respectively. Since the lorry finally comes to rest then the final velocity is zero. Substituting the given figures then
Change in momentum= 6000(0-0.5)=-3000 kg.m/s
Answer:
As the person moves down the zip wire, her increase in kinetic energy is less than her decrease in gravitational potential energy.
Explanation:
Work is done against the air resistance, causing thermal energy to transfer to the surroundings
