1) A toy car is connected by a spring of spring constant k to a block of mass m. The car is parked on a slope with angle θ. (Both car and mass are at rest.
(i) Draw the free body diagrams of the block
(ii) What is the magnitude of the normal force of the ramp on the block?
(iii) What is the magnitude of the force of the spring on the block?
(iv) By what length ∆`p (p is for parked) is the spring stretched from its equilibrium length?
(v) Now let the car accelerate up the ramp with magnitude a, pulling the block after it with the same acceleration. By what length ∆`a (a is for accelerating) is the spring stretched now?
(i) To draw the free body diagram of the block, you need to consider all the forces acting on it.
First, draw the block as a rectangle or a square. Label it with the mass \( m \) and indicate the direction of motion if provided.
The forces acting on the block are:
1. The gravitational force \( mg \) acting downward, where \( m \) is the mass of the block and \( g \) is the acceleration due to gravity (approximately 9.8 m/s\(^2\)).
2. The normal force \( N \) exerted by the ramp on the block. This force acts perpendicular to the surface of the ramp and counterbalances the force of gravity.
3. The force of friction \( f_s \), if there is any due to the roughness of the surface. This force acts parallel to the surface and opposes the motion.
(ii) The magnitude of the normal force of the ramp on the block can be determined using the concept of equilibrium. Since the block is at rest, the sum of the forces in the vertical direction must be zero. Hence, the magnitude of the normal force \( N \) is equal to the magnitude of the gravitational force \( mg \).
(iii) The magnitude of the force of the spring on the block is given by Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position and the spring constant \( k \). Therefore, the magnitude of the force of the spring \( F_{\text{spring}} \) can be calculated as \( F_{\text{spring}} = k \cdot \Delta x \), where \( \Delta x \) is the displacement of the spring from its equilibrium length.
(iv) When the car is parked on the slope, the spring is not stretched, so \( \Delta'p = 0 \).
(v) When the car accelerates up the ramp with magnitude \( a \), it also pulls the block with the same acceleration. In this case, the spring is stretched. To calculate the length \( \Delta'a \) by which the spring is stretched, you need to know the spring constant \( k \) and the displacement \( \Delta'x \) of the spring from its equilibrium position.