Calculating the force due to friction

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The discussion focuses on calculating the force due to friction and involves momentum and gravitational force equations for a man and a sled. The total momentum of the man and sled is calculated as 630, while their gravitational forces are 882 and 980, respectively, totaling 1862. It emphasizes using conservation of momentum to find initial velocity, acceleration, and time based on known variables. A key point raised is the need for clarity on whether to assume constant force, as average force calculations require time information. The conversation also addresses misconceptions regarding the relationship between energy, distance, and average force in friction scenarios.
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Homework Statement
A man (mass is 90.0 kg, 7.00 m/s east) runs towards a sled (mass is 100.0 kg, initially stationary) and jumps on to it (assume momentum is conserved). The sled (with the Agent on top of it) slides a total distance of 4.00 meters. Assume the sled only moves horizontally. Calculate the force due to friction acting on the sled.
Relevant Equations
So I know the formula for friction is Ff= U (Fn) but I have the normal force but I don't have U. I know I can get U but for that I need acceleration. I know I can get acceleration but I need time. But without sounding redundant am I missing something???
p man= mv
p man = 90 (7)
p man= 630

p sled= mv
p sled= 100 (0)
p sled= 0

p man+sled= 630 + 0
p man+sled= 630

fg man= mg
fg man= 90 (9.8)
fg man= 882

fg sled= mg
fg sled= 100 (9.8)
fg sled= 980

fg sled+man= 980+882
fg sled+man= 1862
 

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You should be able to calculate the initial velocity of the (man+sled) by using conservation of momentum. Since you then know the initial velocity and the distance traveled (two variables), you should be able to solve for the acceleration and the time (two variables).
 
phyzguy said:
You should be able to calculate the initial velocity of the (man+sled) by using conservation of momentum. Since you then know the initial velocity and the distance traveled (two variables), you should be able to solve for the acceleration and the time (two variables).
Thank you got what I was looking for really appreciate it.
 
Zero said:
Thank you got what I was looking for really appreciate it.
One quibble.. the question ought to have said to assume the force is constant. Some questions like this ask for average force instead of stating that, but that doesn’t work; for average force you would need to be told the time. Average force x distance does not necessarily equal work done.
 
haruspex said:
One quibble.. the question ought to have said to assume the force is constant. Some questions like this ask for average force instead of stating that, but that doesn’t work; for average force you would need to be told the time. Average force x distance does not necessarily equal work done.
It specifically says the force due to friction. The force due to friction depends only on the normal force, which doesn't change, so it should be constant.
 
phyzguy said:
It specifically says the force due to friction. The force due to friction depends only on the normal force, which doesn't change, so it should be constant.
Coefficients of friction vary, at least a bit, across a surface.
But my reason for intervening is to ensure the OP is not left with the erroneous understanding that energy/distance is a valid way to find average force. It is a widespread misconception.
 
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