Explanation:
The work done by an object is given by :
Here,
F is force
d is displacement
is the angel between F and d.
If the angle between force and displacement is 0, then the work done is equal to, 
.
So, a force does work on an object if a component of the force is parallel to the displacement of the object.
Answer:
Explanation:
From the question we are told that:
Coefficient of kinetic friction 
Vertical Mass 
Horizontal mass 
Generally the equation for kinetic force 
is mathematically given by
Generally the equation for T is mathematically given by
For 
Therefore substituting
Answer: 70 N to the East
Explanation:
lets assume east is positive and west is negative, since they are in opposite directions the net external force = F1+F2
Net force = (-60) + 130
Net force = 70
or
Net force = 70 N in the east direction
Answer:
f = 130 Khz
Explanation:
In a circuit driven by a sinusoidal voltage source, there exists a fixed relationship between the amplitudes of the current and the voltage through any circuit element, at any time.
For an inductor, this relationship can be expressed as follows:
VL = IL * XL (1) , which is a generalized form of Ohm's Law.
XL is called the inductive reactance, and is defined as follows:
XL = ω*L = 2*π*f*L, where f is the frequency of the sinusoidal source (in Hz) and L is the value of the inductance, in H.
Replacing in (1), by the values given of VL, IL, and L, we can solve for f, as follows:
f = VL / 2*π*IL*L = 12 V / 2*π*(3.00*10⁻³) A* (4.9*10⁻³) H = 130 Khz
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>
