A toy car starts at 2.97m/s across a flat surface and encounters friction of -.03328N. How far did it go and how long did it take to stop? The mass of the car is .1132kg Including the formulas would be very helpful. Thanks!
-
Use Newton's second law to find the deceleration caused by friction:
F = m × a
(-0.03328 N) = (0.1132 kg) × a
a = -0.294 m/s²
Now use kinematics formula's to find time and distance.
v = u + a×t
(0 m/s) = (2.97 m/s) + (-0.294 m/s²)×t
t = 10 s < - - - - - - - - - - - - - - - - - - - - - - - answer time
v² = u² + 2×a×s
(0 m/s)² = (2.97 m/s)² + 2×(-0.294 m/s²)×s
s = 15 m < - - - - - - - - - - - - - - - - - - - - - - answer distance
F = m × a
(-0.03328 N) = (0.1132 kg) × a
a = -0.294 m/s²
Now use kinematics formula's to find time and distance.
v = u + a×t
(0 m/s) = (2.97 m/s) + (-0.294 m/s²)×t
t = 10 s < - - - - - - - - - - - - - - - - - - - - - - - answer time
v² = u² + 2×a×s
(0 m/s)² = (2.97 m/s)² + 2×(-0.294 m/s²)×s
s = 15 m < - - - - - - - - - - - - - - - - - - - - - - answer distance
-
F = ma
-0.3328N = 1132 x a
a = -
v = u + at
0 = 2.97 + at
t = -2.97/a
you know a ,so find t
-0.3328N = 1132 x a
a = -
v = u + at
0 = 2.97 + at
t = -2.97/a
you know a ,so find t