Question

A block having a mass of 2 kg projected up 30º incline with initial velocity of 22 m/s. Co-efficient of friction is 0.3. a) How long does the block move up the plane? B) How far does the block move up the plane?

Answer 1

To solve the problem, we need to consider the forces acting on the block and use Newton's laws of motion. Let's break down the forces involved:

1. Weight (mg): The weight of the block acts vertically downward and can be calculated as the product of mass (m) and acceleration due to gravity (g). In this case, the weight is given by W = 2 kg × 9.8 m/s² = 19.6 N.

2. Normal force (N): The normal force acts perpendicular to the incline and opposes the vertical component of the weight. It can be calculated as N = mg cos(θ), where θ is the angle of the incline. In this case, θ = 30º, so N = 2 kg × 9.8 m/s² × cos(30º) = 16.96 N.

3. Frictional force (f): The frictional force acts parallel to the incline and opposes the motion. It can be calculated as f = μN, where μ is the coefficient of friction. In this case, μ = 0.3, so f = 0.3 × 16.96 N = 5.088 N.

4. Component of weight parallel to the incline (W_parallel): This component contributes to the acceleration of the block along the incline and can be calculated as W_parallel = mg sin(θ). In this case, W_parallel = 2 kg × 9.8 m/s² × sin(30º) = 9.8 N.

Now, let's address the questions:

a) How long does the block move up the plane?

To determine the time taken, we need to calculate the acceleration along the incline. The net force acting on the block along the incline can be found by subtracting the frictional force from the component of weight parallel to the incline:

Net force (F_net) = W_parallel - f = 9.8 N - 5.088 N = 4.712 N.

Using Newton's second law of motion (F = ma), we can find the acceleration (a) of the block:

F_net = ma, 4.712 N = 2 kg × a, a = 2.356 m/s².

Next, we can use the kinematic equation to find the time (t) taken to move up the plane:

v = u + at, 0 = 22 m/s + (-2.356 m/s²) × t, t = 22 m/s / 2.356 m/s² = 9.34 s.

Therefore, the block moves up the plane for approximately 9.34 seconds.

b) How far does the block move up the plane?

To determine the distance traveled up the plane, we can use the kinematic equation:

s = ut + (1/2)at²,

where s is the distance, u is the initial velocity, t is the time, and a is the acceleration.

Plugging in the given values:

s = (22 m/s) × (9.34 s) + (1/2) × (-2.356 m/s²) × (9.34 s)²,

s = 205.48 m + (-103.49 m) = 101.99 m.

Therefore, the block moves approximately 101.99 meters up the incline.