Solve for the force of friction acting on a 50 kg carton sliding at 4 m/s if it stops in 3 seconds.?

Answer 1

F = ma, actually. Thanks, ongnoia, I corrected it... :c( I hate dumb mistakes...

Initial velocity = 4
Final velocity = 0
Time = 3 seconds

v = v0 + at
0 = 4 + a(3)
-4 = 3a
a = -4/3 m/s²

So the acceleration due to friction was -4/3 m/s², and the mass is 50kg, so the force should be:

(-4/3 m/s²)(50 kg) = -66.7 N

(The negative meaning that it's against the direction of motion.)

Answer 2

Assuming the speed decreases linearly, the equation for speed would be :
v = at + v0, where a is deceleration/acceleration and v0 the initial velocity,
so :
0 = a x 3 + 4, hence a = -4/3, a negative acceleration, i.e. a deceleration

hence using F = ma, the friction force to stop linearly the movement would amount to : F = - 50 x 4/3 Newton. It's negative as should be expected.

Answer 3

Isn't it F = md?

So 50kg*4m/s*3s= 600kg*m