A block with a mass m = 3.50 kg sits on a surface with a coefficient of static
friction µs = 0.520. A spring with spring constant k = 75.0 N/m is hooked onto the
block, and is slowly pulled horizontally, gradually stretching the spring. (a) When
the spring is stretched 10.0 cm, the block is still stationary. What is the friction force
felt by the block when the spring is stretched 10.0 cm? (b) How much is the spring
stretched when the block just begins to move?
I've done the first part but i'm having trouble knowing what formula to use for the second one. I know you have to use the coefficient of static friction. So i used it to calculate the normal Reaction, but i'm still stuck.
Formulas i have:
F=uN, F=mg, F=ke
Thanks in advance
part (b)
weight of block = normal force = m g
= 3.5 * 9.81 = 34.3 Newtons
maximum static friction force = .52*34.3
= 17.85 N
That is the spring force you need
17.85 = k x
17.85 = 75 x
x = .238 meters = 23.8 cm
To find the spring stretch when the block just begins to move, you need to consider the equilibrium condition. At this point, the force exerted by the spring is equal and opposite to the maximum static friction force acting on the block. Let's break down the solution step by step:
Step 1: Calculate the normal force (N) acting on the block.
Since the block is stationary, the gravitational force (mg) and the normal force (N) must balance each other. Therefore, N = mg, where m is the mass of the block and g is the acceleration due to gravity. In this case, m = 3.50 kg.
Step 2: Calculate the maximum static friction force (Ffmax).
The maximum static friction force can be calculated using the formula Ffmax = µsN, where µs is the coefficient of static friction. In this case, µs = 0.520, and we already have the value of N from the previous step.
Step 3: Equate the maximum static friction force (Ffmax) with the force exerted by the spring (Fspring).
At the point when the block just begins to move, the force exerted by the spring is equal to Fspring = kΔx, where k is the spring constant and Δx is the stretch of the spring. At this point, Fspring = Ffmax.
Step 4: Solve for the spring stretch (Δx) when the block just begins to move.
Set up the equation Ffmax = kΔx and plug in the values:
µsN = kΔx
Now, substitute the known values:
0.520mg = kΔx
Simplify the equation by substituting the values of m, g, and k:
(0.520)(3.50 kg)(9.81 m/s^2) = (75.0 N/m)Δx
Solve for Δx:
Δx = (0.520)(3.50 kg)(9.81 m/s^2) / (75.0 N/m)
By calculating this expression, you will find the value of Δx, which represents the spring stretch when the block just begins to move.