Starting from rest, a body slides down a $45^\circ$ inclined plane in twice the time it takes to slide down the same distance in the absence of friction. The coefficient of friction between the body and the inclined plane is

Question

A rocket of initial mass 6000 kg ejects gases at a constant rate of 16 kg/s with a constant relative speed of 11 km/s. What is the acceleration of the rocket one minute after the blast?

Answer 1

To solve this problem, we need to compare the time taken to slide down an inclined plane with and without friction. We use the basic principles of physics related to motion on an inclined plane, including concepts like gravitational acceleration, friction, and Newton's laws of motion.

First, let's use the following variables:

$ \theta = 45^\circ $
Let $ t_1 $ be the time taken without friction and $ t_2 $ be the time taken with friction. Given: $ t_2 = 2t_1 $
Let $ \mu $ be the coefficient of friction, which we need to calculate.

1. **Time without friction**: Without friction, the only force acting down the plane is the component of gravitational force: F = mg \sin \theta.

From Newton's second law, acceleration $ a_1 $ without friction is given by:

a_1 = g \sin \theta

Using the kinematic equation for starting from rest and moving down distance $ s $:

s = \frac{1}{2} a_1 t_1^2

2. **Time with friction**: With friction, the net force acting down the incline is: F = mg \sin \theta - \mu mg \cos \theta.

The acceleration $ a_2 $ with friction:

a_2 = g \sin \theta - \mu g \cos \theta

Using the kinematic formula:

s = \frac{1}{2} a_2 t_2^2

3. **Relating the equations**: Since the distance $ s $ is the same for both cases, equate the expressions for $ s $:

\frac{1}{2} g \sin \theta t_1^2 = \frac{1}{2} (g \sin \theta - \mu g \cos \theta) (2t_1)^2

4. **Simplifying the equation**:

\sin \theta t_1^2 = (4\sin \theta - 4\mu \cos \theta) t_1^2
Divide both sides by $ t_1^2 $: \sin \theta = 4\sin \theta - 4\mu \cos \theta
3\sin \theta = 4\mu \cos \theta
\Rightarrow \mu = \frac{3\sin \theta}{4\cos \theta}

5. **Calculating $ \mu $**:

For $ \theta = 45^\circ $, $ \sin \theta = \cos \theta = \frac{\sqrt{2}}{2} $:

\mu = \frac{3 \frac{\sqrt{2}}{2}}{4 \frac{\sqrt{2}}{2}} = \frac{3}{4} = 0.75

Therefore, the coefficient of friction is 0.75.

The correct answer is 0.75.

Answer 2