The answer is B tell me if I am wrong.
The object does not move.
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:
a. 120 W
b. 28.8 N
Explanation:
To a good approximate, the only external force that does work on a cyclist moving on level ground is the force of air resistance. Suppose a cyclist is traveling at 15 km/h on level ground. Assume he is using 480 W of metabolic power.
a. Estimate the amount of power he uses for forward motion.
b. How much force must he exert to overcome the force of air resistance?
(a)
He is 25% efficient, therefore the cyclist will be expending 25% of his power to drive the bicycle forward
Power = efficiency X metabolic power
= 0.25 X 480
= 120 W
(b)
power if force times the velocity
P = Fv
convert 15 km/h to m/s
v = 15 kmph = 4.166 m/s
F = P/v
= 120/4.166
= 28.8 N
definition of terms
power is the rate at which work is done
force is that which changes a body's state of rest or uniform motion in a straight line
velocity is the change in displacement per unit time.
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