In a tennis serve, a 0.070 kg ball can be accelerated from rest to 35 m/sec over a distance of 0.80 m. find the magnitude of the average force exerted by the racket on the ball during the serve
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To find the magnitude of the average force exerted by the racket on the ball during the serve, we can use Newton's second law of motion, which states that force (F) is equal to mass (m) multiplied by acceleration (a): F = m * a.
In this case, the given mass (m) of the tennis ball is 0.070 kg, and the acceleration (a) can be found using the following kinematic equation:
v^2 = u^2 + 2 * a * s
Where:
v = final velocity = 35 m/s
u = initial velocity = 0 m/s (since the ball starts from rest)
a = acceleration (to be calculated)
s = distance travelled = 0.80 m
Rearranging the equation to solve for acceleration (a):
a = (v^2 - u^2) / (2 * s)
Plugging in the values:
a = (35^2 - 0^2) / (2 * 0.80)
Calculating:
a = 1225 / 1.60
a = 765.625 m/s^2
Now, we can find the magnitude of the average force (F) exerted by the racket on the ball during the serve:
F = m * a
F = 0.070 kg * 765.625 m/s^2
Calculating:
F = 53.59375 N
Therefore, the magnitude of the average force exerted by the racket on the ball during the serve is approximately 53.6 Newtons (N).