velocity tiime graph of a block of mass 100g sliding on a horizontal concrete floor under the action of a constant force of 5N is shown below. the magnitude of frictional force acting on a block due to floor is ?
A is correct
To determine the magnitude of the frictional force acting on the block, we need to analyze the given velocity-time graph.
The graph shows that the slope of the line is constant, which means that the velocity is changing at a constant rate.
From the graph, we can see that the initial velocity is zero (at time t=0) and the final velocity is 2 m/s (at time t=10 seconds).
We know that acceleration is the rate of change of velocity. Since the acceleration is constant in this case, we can use the formula:
a = Δv / Δt,
where 'a' is the acceleration, Δv is the change in velocity, and Δt is the change in time.
We can calculate the acceleration as follows:
a = (2 m/s - 0 m/s) / (10 seconds - 0 seconds) = 0.2 m/s^2.
Now, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration:
Fnet = m * a,
where 'Fnet' is the net force, 'm' is the mass of the block, and 'a' is the acceleration.
Given that the mass of the block is 100g (or 0.1 kg) and the acceleration is 0.2 m/s^2, we can calculate the net force:
Fnet = 0.1 kg * 0.2 m/s^2 = 0.02 N.
Since the constant force acting on the block is 5N, and the net force is equal to the sum of all the forces acting on the block, we can write:
Fnet = 5 N - frictional force.
Solving for the frictional force:
frictional force = 5 N - 0.02 N = 4.98 N.
Therefore, the magnitude of the frictional force acting on the block due to the floor is approximately 4.98 N.