1. how much force is needed to accelerate a 60 kg skier at 2 m/sec²?\n2. what is the force on a 1000 kg…
1. how much force is needed to accelerate a 60 kg skier at 2 m/sec²?\n2. what is the force on a 1000 kg elevator that is falling freely at 9.8 m/sec²?\n3. what is the acceleration of a 50 kg object pushed with a force of 200 n?\n4. the mass of a large car is 1000 kg. how much force would be required to accelerate the car at a rate of 3 m/sec²?\n5. a 50 kg skater pushed by a friend accelerates 5 m/sec². how much force did the friend apply?\n6. a force of 250 n is applied to an object that accelerates at a rate of 5 m/sec². what is the mass of the object?\n7. a bowling ball rolled with a force of 15 n accelerates at a rate of 3 m/sec²; a second ball rolled with the same force accelerates 4 m/sec². what are the masses of the two balls?\n8. if a 60 kg person on a 15 kg sled is pushed with a force of 300 n, what will be person’s acceleration?\n9. a force of 20 n acts upon a 5 kg block. calculate the acceleration of the object.\n10. an object of mass 300 kg is observed to accelerate at the rate of 4 m/s². calculate the force required to produce this acceleration.\n11. a 5 kg block is pulled across a table by a horizontal force of 40 n with a frictional force of 8 n opposing the motion. calculate the acceleration of the object.\n12. an object of mass 30 kg is in free - fall in a vacuum where there is no air resistance. determine the acceleration of the object.\n13. an object of mass 30 kg is falling in air and experiences a force due to air resistance of 50 newtons.\n a. determine the net force acting on the object and\n b. calculate the acceleration of the object.
Answer
Explanation:
Step1: Recall Newton's second - law
The formula is $F = ma$, where $F$ is force in Newtons (N), $m$ is mass in kilograms (kg), and $a$ is acceleration in meters per second squared ($m/s^{2}$). We will use this formula to solve each problem.
Problem 1:
Given $m = 60$ kg and $a= 2$ $m/s^{2}$, find $F$. $F=ma=60\times2 = 120$ N
Problem 2:
Given $m = 1000$ kg and $a = 9.8$ $m/s^{2}$ (acceleration due to gravity for free - fall near Earth's surface), $F=ma=1000\times9.8=9800$ N
Problem 3:
Given $F = 500$ N and $m = 50$ kg, find $a$. $a=\frac{F}{m}=\frac{500}{50}=10$ $m/s^{2}$
Problem 4:
Given $m = 1000$ kg and $a = 3$ $m/s^{2}$, $F=ma=1000\times3 = 3000$ N
Problem 5:
Given $m = 50$ kg and $a = 5$ $m/s^{2}$, $F=ma=50\times5=250$ N
Problem 6:
Given $F = 250$ N and $a = 5$ $m/s^{2}$, $m=\frac{F}{a}=\frac{250}{5}=50$ kg
Problem 7:
For the first ball: Given $F = 15$ N and $a = 3$ $m/s^{2}$, $m_1=\frac{F}{a}=\frac{15}{3}=5$ kg For the second ball: Given $F = 15$ N and $a = 4$ $m/s^{2}$, $m_2=\frac{F}{a}=\frac{15}{4}=3.75$ kg
Problem 8:
The total mass $m=m_{person}+m_{sled}=60 + 15=75$ kg, and $F = 300$ N. $a=\frac{F}{m}=\frac{300}{75}=4$ $m/s^{2}$
Problem 9:
Given $F = 20$ N and $m = 5$ kg, $a=\frac{F}{m}=\frac{20}{5}=4$ $m/s^{2}$
Problem 10:
Given $m = 300$ kg and $a = 4$ $m/s^{2}$, $F=ma=300\times4 = 1200$ N
Problem 11:
The net force $F_{net}=F - f$, where $F = 40$ N and $f = 8$ N, so $F_{net}=40 - 8=32$ N. Given $m = 5$ kg, $a=\frac{F_{net}}{m}=\frac{32}{5}=6.4$ $m/s^{2}$
Problem 12:
In free - fall in a vacuum, the acceleration of an object is equal to the acceleration due to gravity, $a = 9.8$ $m/s^{2}$
Problem 13a:
The force due to gravity $F_g=mg$, where $m = 30$ kg and $g = 9.8$ N/kg, so $F_g=30\times9.8 = 294$ N. The air resistance $F_{air}=50$ N. The net force $F_{net}=F_g - F_{air}=294-50 = 244$ N
Problem 13b:
Given $m = 30$ kg and $F_{net}=244$ N, $a=\frac{F_{net}}{m}=\frac{244}{30}\approx8.13$ $m/s^{2}$
Answer:
- 120 N
- 9800 N
- 10 $m/s^{2}$
- 3000 N
- 250 N
- 50 kg
- 5 kg, 3.75 kg
- 4 $m/s^{2}$
- 4 $m/s^{2}$
- 1200 N
- 6.4 $m/s^{2}$
- 9.8 $m/s^{2}$ 13a. 244 N 13b. 8.13 $m/s^{2}$