A stone, of mass 700g......?

Question

a stone, of mass 700g is thrown with a velocity of 20 ms^-2 across the frozen surface of a pond and comes to rest after traveling 40m. What is the force of friction between the stone and ice?

i thought it is:

f=ma

f=0.7xa

v=u+at

a=(v-u)/t

f=0.7x(v-u)/t

f=0.7x(20)/4

f=0.7x5

f=3.5N?

Answer 1

First calculate the acceleration of the staone

V² = U² +2*a*s

0= 20²+2*40*a

0 = 400+80a

a = - 400/80

a = -5m/s² -ve sign comfirms stone is slowing down.

Force = m*a

Force = 0.700 * -5

Force = -3.5N  -ve sign confirms that the force is acting opposite to the direction of the stone, ie it is a friction force.

Answer: The friction force between the stone and the ice is 3.5N

Answer 2

Don't forget when u throw a body,acceleration due to gravity(g) is taking part an also velocity is m/s not m/s^2.Force=frict force+mass*acce due grav

Force=mg=0.7*10=7N

Am still trying solving it...

Answer 3

Sorry, I can't help, but how nice for someone to actually show their own thoughts and workings, rather than just putting the homework question and getting others to do it. Well done!

Answer 4

the answer is right, but i dont see how u got there...

knowing m=0.7kg u=20ms^-1, s=40m  v=0ms^-1

first find a

v^2 = u^2 + 2as

0 = 400 +80a

-80a = 400

a = -5ms^-2 in ---> direction

therefore is 5ms^-2 in <---- direction  (opposit direction to movement)

so friction is acting against the movement

F=ma

= 0.7 x 5

= 3.5N