A small bob of mass m=.01kg is hanging from the roof of a moving car in equilibrium making angle theta with vertical. If initial velocity of car was zero and acceleration of car is 2ms^-2.find work sink done by tension in ground frame in first 10 seconds.?

Question

Answer

**Introduction:**
In this problem, we have a small bob hanging from the roof of a moving car in equilibrium making an angle theta with the vertical. We need to find the work done by the tension in the ground frame in the first 10 seconds, given that the initial velocity of the car was zero and the acceleration of the car is 2ms^-2.

**Solution:**

**Step 1: Resolving forces**
The forces acting on the bob are the tension T and the weight mg. Resolving the forces in the vertical and horizontal direction, we get:

In the vertical direction, Tcos(theta) - mg = 0 ---------(1)
In the horizontal direction, Tsin(theta) = ma ----------(2)

**Step 2: Finding the tension**
From equation (1), we get:
Tcos(theta) = mg
T = mg/cos(theta)

Substituting this value of T in equation (2), we get:
Tsin(theta) = ma
mg tan(theta) = ma
a = g tan(theta)

**Step 3: Finding the displacement**
The displacement of the car in the first 10 seconds can be calculated using the formula:
s = ut + 1/2 at^2
where u = 0, a = 2ms^-2, and t = 10s

s = 0 + 1/2 x 2 x 10^2
s = 100m

**Step 4: Finding the work done**
The work done by the tension is given by the formula:
W = Fs
where F is the force and s is the displacement.

The tension T does work on the bob by lifting it up against the force of gravity. The work done by the tension in lifting the bob through a vertical height h is given by:
W = mgh

Using the value of T from step 2 and the value of h, we get:
W = mg/cos(theta) x h
h = s sin(theta)
h = 100sin(theta)

Substituting the value of h in the above equation, we get:
W = mg/cos(theta) x 100sin(theta)
W = 1.96 x 10^-3 Joules

Therefore, the work done by the tension in the ground frame in the first 10 seconds is 1.96 x 10^-3 Joules.

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