A 10 kg block sits on a flat surface whose μs=0.60 and whose μk is 0.40
a. What horizontal force is required to get the block move?
b.If we continue to apply the same force as in part a. what will the blocks acceleration be?
To find the horizontal force required to get the block to move, you need to consider the static friction. The static friction opposes the force applied to the block until it reaches its maximum value, which is given by the equation:
fs ≤ μs * N
Where:
fs is the static friction force
μs is the coefficient of static friction
N is the normal force
The normal force, N, can be calculated as the weight of the block since it is sitting on a flat surface. The weight of an object is given by the equation:
weight = mass * acceleration due to gravity
So, the weight of the block is:
weight = 10 kg * 9.8 m/s^2
weight = 98 N
The maximum static friction force is:
fs ≤ 0.60 * 98 N
fs ≤ 58.8 N
Therefore, a horizontal force of at least 58.8 Newtons is required to get the block to move.
For part b, once the block starts moving, the force of static friction is replaced by the force of kinetic (sliding) friction. The kinetic friction force is determined using the equation:
fk = μk * N
Where:
fk is the kinetic friction force
μk is the coefficient of kinetic friction
N is the normal force (same as before)
Using the same normal force value as before (98 N), the kinetic friction force is:
fk = 0.40 * 98 N
fk = 39.2 N
Since there is a constant force of 39.2 N acting on a 10 kg block, you can calculate the acceleration using Newton's second law of motion:
F = ma
Rearranging the equation, you get:
a = F / m
Substituting the values, you get:
a = 39.2 N / 10 kg
a = 3.92 m/s^2
Therefore, the block will accelerate at a rate of 3.92 m/s^2 when the same force is applied.
force=.6*mg to move
net force=ma
first force-.6mg=ma solve for a