A 970-kg car starts from rest on a horizontal roadway and accelerates eastward for 5.00 s when it reaches a speed of 25.0 m/s. What is the average force exerted on the car during this time
4850N
lkxm
4.85j
To find the average force exerted on the car during this time, we can use Newton's second law of motion, which states that the force applied to an object is equal to its mass multiplied by its acceleration.
Given:
Mass of the car (m) = 970 kg
Initial velocity (u) = 0 m/s
Final velocity (v) = 25.0 m/s
Time taken (t) = 5.00 s
First, we need to calculate the acceleration of the car using the formula:
acceleration (a) = (final velocity - initial velocity) / time taken
Substituting the given values:
a = (25.0 m/s - 0 m/s) / 5.00 s
a = 25.0 m/s / 5.00 s
a = 5.00 m/s²
Now, we can calculate the average force using the formula:
force (F) = mass * acceleration
Substituting the given values:
F = 970 kg * 5.00 m/s²
F = 4850 N
Therefore, the average force exerted on the car during this time is 4850 Newtons.