Question

instructions: use the formula for force to solve the following problems. round all answers to the correct number of significant figures. show all your work.

mass force acceleration calculations

1 365 kg 1.25 m/s²
2 194 kg 1.3 m/s²
3 55.47 kg 81 m/s²
calculate the force exerted by a 52.5 kg cheetah when he accelerates at 9.86 m/s² chasing a gazelle.
4 mass force acceleration calculations
calculate the force a pro tennis player must apply to a 59.4 g tennis ball in order to achieve 435 m/s² of acceleration.
5 mass force acceleration calculations
calculate the force required to accelerate a 1,237 kg car 4.45 m/s² if it must overcome 750 n of friction.
6 mass force acceleration calculations
checksum 1199.8 the checksum is the unrounded total of all correctly calculated and rounded answers in the shaded boxes.

Explanation:

Step1: Recall force formula

The formula for force is $F = m\times a$, where $F$ is force in Newtons (N), $m$ is mass in kilograms (kg), and $a$ is acceleration in meters - per - second squared ($m/s^{2}$).

Step2: Solve problem 1

Given $m = 365$ kg and $a=1.25$ $m/s^{2}$. Then $F=m\times a=365\times1.25 = 456.25$ N. Rounding to 3 significant figures, $F = 456$ N.

Step3: Solve problem 2

Given $m = 194$ kg and $a = 1.3$ $m/s^{2}$. Then $F=m\times a=194\times1.3=252.2$ N. Rounding to 2 significant figures, $F = 250$ N.

Step4: Solve problem 3

Given $m = 55.47$ kg and $a = 81$ $m/s^{2}$. Then $F=m\times a=55.47\times81 = 4493.07$ N. Rounding to 2 significant figures, $F = 4500$ N.

Step5: Solve problem 4

Given $m = 52.5$ kg and $a = 9.86$ $m/s^{2}$. Then $F=m\times a=52.5\times9.86 = 517.65$ N. Rounding to 3 significant figures, $F = 518$ N.

Step6: Solve problem 5

First, convert the mass of the tennis - ball to kg. Since $1$ g=$0.001$ kg, $m = 59.4$ g=$0.0594$ kg and $a = 435$ $m/s^{2}$. Then $F=m\times a=0.0594\times435 = 25.839$ N. Rounding to 3 significant figures, $F = 25.8$ N.

Step7: Solve problem 6

First, find the force required to accelerate the car without considering friction using $F_{1}=m\times a$. Given $m = 1237$ kg and $a = 4.45$ $m/s^{2}$, so $F_{1}=1237\times4.45=5494.65$ N. Then, to overcome a friction force of $F_{f}=750$ N, the total force $F = F_{1}+F_{f}=5494.65 + 750=6244.65$ N. Rounding to 3 significant figures, $F = 6240$ N.

Answer:

#	Mass	Force	Acceleration	Calculations
2	194 kg	250 N	1.3 m/s²	$F = 194\times1.3 = 252.2\approx250$ N
3	55.47 kg	4500 N	81 m/s²	$F = 55.47\times81=4493.07\approx4500$ N
4	52.5 kg	518 N	9.86 m/s²	$F = 52.5\times9.86 = 517.65\approx518$ N
5	0.0594 kg	25.8 N	435 m/s²	$F = 0.0594\times435=25.839\approx25.8$ N
6	1237 kg	6240 N	4.45 m/s²	$F=(1237\times4.45)+750=5494.65 + 750=6244.65\approx6240$ N