a sled mass of 50kg is pulled horizontally over flat ground. the static friction coefficient is .30, and the sliding friction coefficient is .10.
what minimum amount of force must be applied to the sled in order to start it moving? i got 147.15N
what amount of applied force will keep it moving at a constant velocity of 3.0m/s?
To find the minimum amount of force required to start the sled moving, you can use the equation:
F = μ * N
Where:
F is the force needed to overcome friction
μ is the coefficient of static friction
N is the normal force
The normal force can be calculated as the product of the sled's mass (50kg) and the acceleration due to gravity (9.8 m/s²):
N = m * g
N = 50kg * 9.8 m/s²
N = 490 N
Now, substitute the values into the equation:
F = 0.30 * 490 N
F = 147 N
Therefore, the minimum amount of force required to start the sled moving is 147 N.
To find the amount of force required to keep the sled moving at a constant velocity of 3.0 m/s, you need to consider the equation:
F = μ * N
But this time, you need to use the coefficient of sliding friction (μ = 0.10) since the sled is already in motion. Additionally, there is no mention of any other external forces acting on the sled, so we can assume that the force required is equal to the force of sliding friction.
First, calculate the normal force:
N = m * g
N = 50kg * 9.8 m/s²
N = 490 N
Next, substitute the values into the equation:
F = 0.10 * 490 N
F = 49 N
Therefore, the amount of applied force required to keep the sled moving at a constant velocity of 3.0 m/s is 49 N.
a. M*g = 50 * 9.8 = 490 N. = Wt. of sled = Normal force(Fn).
Fs = us*Fn = 0.3 * 490 = 147 N.
Fap-Fs = M*a.
Fap = M*0 + 147 = 147 N.
b. Fk = uk*Fn = 0.10 * 490 = 49 N.
Fap = M*0 + 49 = 49N.