A block of mass 1.4kg is placed on a rough surface. the coefficient of friction between the surfaces is 0.58. A constance force of magnitude 10N, making an angle of 37 degrees with the horizontal is acting on the block. the block is given an initial velocity of 14 m/s towards right at t=0. find the distance travelled by the block in a time interval of (a)0.5s (b)1.5s
To find the distance traveled by the block in a time interval, we can break down the problem into several steps.
Step 1: Calculate the Net Force
The net force acting on the block can be calculated by resolving the applied force into horizontal and vertical components.
Horizontal component: F_hor = F_applied * cos(θ)
Vertical component: F_ver = F_applied * sin(θ)
Given:
F_applied = 10N
θ = 37 degrees
Therefore,
F_hor = 10N * cos(37)
F_hor ≈ 8N
Step 2: Calculate the Frictional Force
The frictional force can be calculated using the equation:
Frictional force (F_friction) = coefficient of friction * Normal force
The normal force (N) can be calculated as the product of the mass and the acceleration due to gravity (N = m * g).
Given:
m = 1.4kg
coefficient of friction (μ) = 0.58
acceleration due to gravity (g) ≈ 9.8 m/s^2
First, calculate the normal force:
N = m * g
N = 1.4kg * 9.8 m/s^2
N ≈ 13.72N
Then, calculate the frictional force:
F_friction = μ * N
F_friction = 0.58 * 13.72N
F_friction ≈ 7.96N
Step 3: Calculate the Net Horizontal Force
The net horizontal force can be calculated by subtracting the frictional force from the horizontal component of the applied force:
Net horizontal force = F_hor - F_friction
Net horizontal force = 8N - 7.96N
Net horizontal force ≈ 0.04N
Step 4: Calculate the Acceleration
The acceleration of the block can be calculated using Newton's second law:
Net horizontal force = mass * acceleration
Given:
mass (m) = 1.4kg
Therefore:
0.04N = 1.4kg * acceleration
acceleration ≈ 0.03 m/s^2
Step 5: Calculate the Distance Traveled
a) Time interval = 0.5s
Using the equation for distance traveled (d) with constant acceleration:
d = initial velocity * time + (1/2) * acceleration * time^2
Given:
initial velocity = 14 m/s
time interval = 0.5s
Using the above values and the previously calculated acceleration:
d = 14 m/s * 0.5s + (1/2) * 0.03 m/s^2 * (0.5s)^2
d = 7m + 0.01875m
d ≈ 7.02m
Therefore, the block travels approximately 7.02 meters in a time interval of 0.5 seconds.
b) Time interval = 1.5s
Using the same equation for distance:
d = initial velocity * time + (1/2) * acceleration * time^2
Given:
initial velocity = 14 m/s
time interval = 1.5s
Using the above values and the previously calculated acceleration:
d = 14 m/s * 1.5s + (1/2) * 0.03 m/s^2 * (1.5s)^2
d = 21m + 0.03375m
d ≈ 21.03m
Therefore, the block travels approximately 21.03 meters in a time interval of 1.5 seconds.