Can anyone solve this physics problem?

Yes, this is pretty basic. Why don’t you read the book that these problems are in – I’m sure it provides all the info you need to solve these problems in there.

…and if you solve it on your own, the next will might come easier. ;-)

Perhaps we should devise a method in which users get paid for doing other users homework. I’ll help you for $5.

I have enough of my own homework to complete right now.

I have worked it out and I just want to make sure that my answer is correct. Can someone work it out for me to check it….Please…..

How could the kid jump on a frictionless surface?

54 newtons on the sled? A 50kg kid is going to fall OFF that sled, the time at which s/he falls off will be hard to work out without knowing the kid.

6 meters per second is FAST. That’s about 13.5 mph, or a pretty nice clip on a BIKE. If you ran this for a mile, you’d be running 4:26 miles. Usan Bolt does a nice clip around 10m/s, but this is damn FAST for a child.

Physics problems these days.

Anyway, I get 0.7109 BTU (seriously)

I won’t make fun of you.

“Calculate the speed of the sled with the child after she jumps onto the sled.”

Momentum is conserved; the total momentum of the system (child + sled) is the same before and after the child jumps on the sled.

Before the child jumps on the sled, the momentum of the child is (p = mv) 50kg * 6m/s = 300kgm/s, and the momentum of the block is 10kg * 0m/s = 0kgm/s. So the total momentum of the system is 300kgm/s + 0kmg/s = 300kgm/s.

This total momentum is conserved. Before and after the child jumps on the sled, the total momentum must equal 300kgm/s.

After the child jumps on the sled, the momentum is p = (child mass + sled mass) * velocity, or p = (50kg + 10kg) * v = 60 * v.

Using momentum conservation, we set the momentum before the event equal to the momentum afterwards. 300kgm/s = 60 * v —> v = 5m/s

“Calculate the kinetic energy of the sled with the child after she jumps onto the sled.”

Kinetic energy obeys KE = (½)mv^2. Assuming they question is asking for the kinetic energy of the total system (child + sled), simply substitute the known values. KE = (½)(60kg)(5m/s)^2 = 750J.

“How much work must be done by friction to bring the sled with the child to a stop?”

From the Work-Kinetic Energy Theorem, we know that the amount of work done on a body is equal to the change in kinetic energy of the body. The system starts with 750J, and ends with 0J (v = 0), so the change in KE, and thus the work done by friction to stop the system, is 750J. The friction force give here seems to be irrelevant, although you could use it to determine how far the system slides before it stops.

I’ll show you mine if you show me yours.

I just did!

Were we supposed to time ourselves?

Calculate the speed of the sled with the child after she jumps onto the sled.

Take the law of conservation of momentum i.e total momentum before collision = total momentum after collision.
The mass of kid =50 Kg and the mass of sled is 10 kg.After the kid is on the sled the mass all together is 60kg
So momentum before collision =50×6 and that of after collision is 60xv
applying law of conservation of momentum
50*6 = 60*v
=>v= 5mts/sec

Calculate the speed of the sled with the child after she jumps onto the sled.
½ mv^2 =½*60*25 = 750 Joules

After a short time, the moving sled with the child aboard reaches a rough level surface that exerts a constant frictional force of 54 newtons on the sled. How much work must be done by friction to bring the sled with the child to a stop?
Thanks alot!!!

This is a good.Till the time the sled with the kid was on a friction less floor,it’s speed was constant i.e 5m/s which is it’s initial speed.When it came on the friction floor it reduced and finally became zero.
The friction force was the only deceleration actiing on it, so the deceleration or the negetive acceleration was 54/60 = 0.9 mts /sec

Now we have to find out the displacement on rough floor in order to get the work done.
Putting the equation v^2-U^2=2as we get
-(5)^2=-2*0.9*S
=>s=25/1.8 = 13.88 mts
So Work done by friction is Force X displacement =54*13.88 = 750 Joules.