What Force Accelerates a Sled on Snowy Ground?

[mmoran]

Homework Statement

A sled of mass 13.4 kilograms is pulled along flat snow-covered ground. The static friction coefficient is 0.334, and the kinetic friction coefficient is 0.111.
I solved both part a and b, but I am having trouble with part c.
c. Once moving, what force is needed to accelerate the sled 2.69 meters per second per second?

Homework Equations

am=f
u=f/n

The Attempt at a Solution

Im confused on how to do this.
So Its asking for the force pushing the sled forward. Using the formula am=f
if i plug in acceleration and the mass i get 13.4x2.69=36.046, but it said it was wrong.
Am I missing something? Do I need to factor in frictional force?
Would I add 14.6 to 36.046? 14.6 is the force needed to push the object at a constant velocity.

 

[Andrew Mason]

Homework Statement

A sled of mass 13.4 kilograms is pulled along flat snow-covered ground. The static friction coefficient is 0.334, and the kinetic friction coefficient is 0.111.
I solved both part a and b, but I am having trouble with part c.
c. Once moving, what force is needed to accelerate the sled 2.69 meters per second per second?

Homework Equations

am=f
u=f/n

The Attempt at a Solution

Im confused on how to do this.
So Its asking for the force pushing the sled forward. Using the formula am=f
if i plug in acceleration and the mass i get 13.4x2.69=36.046, but it said it was wrong.
Am I missing something? Do I need to factor in frictional force?
Would I add 14.6 to 36.046? 14.6 is the force needed to push the object at a constant velocity.

Do a freebody diagram of the accelerating sleigh. What are the forces acting on the sleigh? What do they necessarily sum to?

AM

 

[kerimek]

Hello,

In this problem, we are given the mass of the sled (13.4 kg) and the acceleration (2.69 m/s^2). We are also given the two coefficients of friction, static (0.334) and kinetic (0.111).

To find the force needed to accelerate the sled, we need to use the formula F=ma, where F is the force, m is the mass, and a is the acceleration. So, in this case, the force needed would be:

F = 13.4 kg x 2.69 m/s^2 = 36.046 N

This is the total force needed to accelerate the sled, but we need to take into account the friction forces as well.

The friction force can be calculated using the formula Ff=μN, where Ff is the friction force, μ is the coefficient of friction, and N is the normal force. The normal force, in this case, is equal to the weight of the sled, which is mg, where g is the acceleration due to gravity (9.8 m/s^2).

So, for the static friction force, we have:

Ff = μN = (0.334)(13.4 kg)(9.8 m/s^2) = 42.136 N

And for the kinetic friction force, we have:

Ff = μN = (0.111)(13.4 kg)(9.8 m/s^2) = 13.076 N

Now, to find the net force needed to accelerate the sled, we need to subtract the friction force from the total force:

Net force = 36.046 N - 42.136 N = -6.09 N

Note that the net force is negative because the friction force is acting in the opposite direction of the motion.

So, the force needed to accelerate the sled 2.69 m/s^2 would be 6.09 N in the direction of motion. I hope this helps! Let me know if you have any other questions.