A 2 kg block is being pulled across a frictionless surface as shown below. The magnitude of the force is 10 Newtons. How much work is done on the box after 5 seconds? (it's number 3 on the website)
http://www.chs.d211.org/science/torpedw/AP%20Physics%20C/Unit%203-Energy/Conservation%20of%20Energy%20Problems.pdf

Question

A 2 kg block is being pulled across a frictionless surface as shown below. The magnitude of the force is 10 Newtons. How much work is done on the box after 5 seconds? (it's number 3 on the website)
http://www.chs.d211.org/science/torpedw/AP%20Physics%20C/Unit%203-Energy/Conservation%20of%20Energy%20Problems.pdf

anonymous · Answer

force in horizontal direction would be F= 10 cos 45 =5 N
so work done W= F.d
also d= at^2 /2 and a=F/m

anonymous · Answer

The answer does not equal 312 J however ):

anonymous · Answer

oh sorry F= 10 cos 45 = 10/ sqrt 2= 5 sqrt 2

anonymous · Answer

It still does not equal 312J, right?

anonymous · Answer

did you get your answer?

anonymous · Answer

Sadly, no. ):

anonymous · Answer

ok
force in horizontal direction would be F= 10 cos 45 =5sqrt 2 N
so work done W= F.d
also d= at^2 /2 and a=F/m=5sqrt 2/2
so 
$$W=F.d =5 \sqrt 2\times \frac{ 5 \sqrt 2 }{ 2\times2} \times5^2 = 312.5 J$$

anonymous · Answer

ohh thank you so much!!!