A stone of 1 kg is thrown with a velocity of 20 m s−1 across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?
Answers
Answer:
a stone of 1 kg is thrown with velocity of 20m/s across the frozen surface of a lake and comes to rest after travelling a ...
m = 1kgu = 20m/s. v = 0m/s. s(distance travelled) = 50musing third equation of motionv²=u²+2as0²
Given:
- Initial velocity,u = 20 m/s
- Final velocity,v = 0 [ stone comes to rest ]
- Distance travelled,s = 50 m
- Mass of object,m = 1 kg
To be calculated:
Calculate the force of friction between the stone and the ice?
Formula used:
- v² = u² + 2as
- F = m × a
Solution:
According to the third equation of motion ,
v² = u² + 2as
★ Substituting the values in the third equation of motion,we get :
⇒ ( 0 )² = ( 20 )² + 2 × a × 50
⇒ 0 = 400 + 100a
⇒ 100a = -400
⇒ a = - 400/100
⇒ a = -4 m/s²
Now,
Force = m × a
★ putting the values of mass and acceleration, we get :
Force = 1 × ( - 4 )
= -4 N
Thus,the force of friction between the stone and the ice is 4 Newtons. [ Note : The negative sign show that this force oppose the motion of stone ]