A 5.0-kg block is at rest on a horizontal floor. If you push horizontally on the 5.0-kg block with a force of 10.0 N, it just starts to move.
(a) What is the coefficient of static friction?
(b) A 12.0-kg block is stacked on top of the 5.0-kg block. What is the magnitude F of the force, acting horizontally on the 5.0-kg block as before, that is required to make the two blocks start to move?
(a) (mu_static) = F/(M*g) = 10/(5*9.8) = ____
(b) The force required to make the stack move will be higher by a factor 17/5 (the weight increase ratio)
(17/5)*10 = ___ N
THANK YOU :)
To find the coefficient of static friction (μs), we can use the equation:
μs = F / N
where F is the force applied horizontally (10.0 N) and N is the normal force. The normal force is equal to the weight of the block, which is given by:
N = m * g
where m is the mass of the block (5.0 kg) and g is the acceleration due to gravity (9.8 m/s^2).
(a) Calculating the coefficient of static friction:
N = 5.0 kg * 9.8 m/s^2 = 49 N
μs = 10.0 N / 49 N ≈ 0.2041
Therefore, the coefficient of static friction is approximately 0.2041.
(b) To find the magnitude of force (F) required to make the two blocks start to move, we need to consider the added weight from the 12.0-kg block. The total mass is now 17.0 kg (5.0 kg + 12.0 kg).
N = (5.0 kg + 12.0 kg) * 9.8 m/s^2 = 166.6 N
Using the same equation as before:
μs = F / N
we can rearrange to solve for F:
F = μs * N
Substituting the known values:
F = 0.2041 * 166.6 N ≈ 34 N
Therefore, the magnitude of the force required to make the two blocks start to move is approximately 34 N.
To answer both parts of the question, we need to calculate the maximum value of static friction, since that is the force that must be overcome in order for the blocks to start moving.
(a) To find the coefficient of static friction, we can use the formula:
F_smax = μ_s * N
where F_smax is the maximum value of static friction, μ_s is the coefficient of static friction, and N is the normal force.
Since the block is on a horizontal floor and at rest, the normal force N is equal to the weight of the block, which is given by:
N = m * g
where m is the mass of the block and g is the acceleration due to gravity.
Given:
Mass of the block (m) = 5.0 kg
Force applied (F) = 10.0 N
Using the value of force (F), we can determine the magnitude of the maximum static friction that you applied to the block, F_s:
F_s = F
Since the block is just starting to move, F_s is equal to the maximum static friction F_smax.
Plugging the values into the formula, we get:
F_smax = F = μ_s * N
Substituting the expression for N = m * g, we have:
F_smax = μ_s * (m * g)
Now, let's solve for μ_s:
μ_s = F_smax / (m * g)
Substituting the given values, we find:
μ_s = 10.0 N / (5.0 kg * 9.8 m/s^2)
μ_s ≈ 0.204
Therefore, the coefficient of static friction is approximately 0.204.
(b) Now let's consider the situation where a 12.0-kg block is stacked on top of the 5.0-kg block.
The normal force N will be the sum of the weights of both blocks:
N = (m1 + m2) * g
= (5.0 kg + 12.0 kg) * 9.8 m/s^2
= 17.0 kg * 9.8 m/s^2
= 166.6 N
Again, we need to calculate the maximum value of static friction that needs to be overcome for the blocks to start moving.
The maximum static friction is given by:
F_smax = μ_s * N
Substituting the values, we find:
F_smax = 0.204 * 166.6 N
≈ 33.99 N
Therefore, in order to make the two blocks start moving horizontally, a force of magnitude approximately 33.99 N needs to be applied to the 5.0-kg block.