a bock with mass 60kg sits on a surface with static friction Us = 0.51. a spring (k = 69N) is stretched horizontally by 11cm but the block is still stationary. What is the friction force felt by the block when stretched 11cm?
To determine the friction force felt by the block when it is stretched by 11 cm, we need to consider the balance of forces acting on the block.
First, let's consider the spring force. The spring force can be calculated using Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. The formula for the spring force is given by:
Fs = -k * x
Where Fs is the spring force, k is the spring constant (69 N/m in this case), and x is the displacement of the spring from its equilibrium position (11 cm or 0.11 m in this case).
Substituting the given values, we get:
Fs = -69 N/m * 0.11 m
Fs = -7.59 N
Now, let's consider the static friction force. The static friction force can be calculated using the formula:
Fs = μs * N
Where Fs is the friction force, μs is the coefficient of static friction, and N is the normal force. The normal force is the force exerted by the surface on the block in a direction perpendicular to the surface.
Since the block is stationary, the net force acting on it must be zero. This means that the friction force must be equal in magnitude but opposite in direction to the spring force, in order to cancel it out. Therefore, we have:
Fs = μs * N
-7.59 N = 0.51 * N
Now, solve for N:
N = -7.59 N / 0.51
N ≈ -14.88 N
Since the normal force cannot be negative, we take the absolute value:
N ≈ 14.88 N
Therefore, the friction force felt by the block when it is stretched 11 cm is approximately 14.88 Newtons.