```Newton’s Second Law Practice
1. A sled of mass 70kg is pushed across a frictionless ice surface with a force of 220N. If it starts from
rest,
a) find the acceleration
๐น = ๐๐
220 = 70๐
2
๐ = 3. 1 ๐/๐
b) find the velocity after 10 seconds.
๐ฃ = ๐ฃ0 + ๐๐ก
๐ฃ = 0 + (3. 1)(10)
๐ฃ = 31 ๐/๐
2. A 561kg rocket has a thrust force of 18500N. It starts from rest on the ground. Find its height after 10
seconds.
๐น = ๐๐ − ๐๐
18500 = 561๐ − 561(9. 8)
2
๐ = 33. 0 ๐/๐
๐๐๐๐ก = 33. 0 − 9. 8
2
๐๐๐๐ก = 23. 2 ๐/๐
โ = ๐ฃ0๐ก +
1
2
2
๐๐ก
โ = (10)(10) +
โ = 0 + 1160
โ = 1160 ๐
1
2
2
(23. 2)(10)
3. A 530kg snowmobile is pushed forwards by a force of 2700N. There is a friction force of 750N and
air resistance of 300N.
a) Find the net force
๐น๐๐๐ก = 2700๐ − 750๐ − 300๐
๐น๐๐๐ก = 2650๐
b) Find the acceleration.
๐น๐๐๐ก = ๐๐
2650 = 530๐
2
๐ = 5 ๐/๐
4. A 40kg sled is pulled with a force of 250N and accelerates at 1.5m/s2.
a) Find the force of friction.
๐น๐๐๐ก = ๐๐
250 − ๐ = (40)(1. 5)
250 − 60 = ๐
๐ = 190๐
b) Find the coefficient of friction
๐ = ๐๐
๐ = (40)(9. 8)
๐ = 392๐
๐
μ = ๐
μ =
190
392
μ = 0. 48
c) Find the acceleration of the sled if the person pulling lets go
๐น๐๐๐ก = ๐๐
− 190 = 40๐
๐ =
2
− 4. 75 ๐/๐
5. a) Find the net force needed to accelerate a 1200 kg car from 20 to 30 km/h in 2.7 seconds
20 ๐๐/โ &times;
30 ๐๐/โ &times;
1000 ๐
1 ๐๐
1000 ๐
1 ๐๐
&times;
&times;
1โ
3600 ๐
1โ
3600 ๐
= 5. 67 ๐/๐
= 8. 33 ๐/๐
๐ฃ = ๐ฃ0 + ๐๐ก
8. 33 = 5. 67 + 2. 7๐
2. 77 = 2. 7๐
2
๐ = 1. 029 ๐/๐
๐น๐๐๐ก = ๐๐
๐น๐๐๐ก = (1200)(1. 029)
๐น๐๐๐ก = 1235๐
b) What is the minimum coefficient of friction which allows this acceleration?
๐ = ๐๐
๐ = (1200)(9. 8)
๐ = 11760๐
๐ = ๐น๐๐๐ก
๐ = 1235๐
๐
μ = ๐
μ =
1235
11760
μ = 0. 10
6. Find the acceleration of a 15 kg sled pulled with a force of 120N if
a) There is no friction.
๐น๐๐๐ก = ๐๐
120 = 15๐
2
๐ = 8 ๐/๐
b) There is a 50N friction force
๐น๐๐๐ก = ๐๐
120 − 50 = 15๐
70 = 15๐
2
๐ = 4. 7 ๐/๐
c) μ =0.3
๐ = ๐๐
๐ = (15)(9. 8)
๐ = 147๐
๐ = μ๐
๐ = (0. 3)(147)
๐ = 44. 1๐
๐น๐๐๐ก = ๐๐
120 − 44. 1 = 15๐
75. 9 = 15๐
2
๐ = 5. 06 ๐/๐
7. A 7000kg jet accelerates forwards against a 27 000N drag force.
a) Find the acceleration if the thrust is 39000 N
๐น๐๐๐ก = ๐๐
39000 − 27000 = 7000๐
12000 = 7000๐
2
๐ = 1. 7 ๐/๐
b) Find the Thrust required to produce an acceleration of 2.7 m/s2
๐น๐๐๐ก = ๐๐
๐น๐ก − 27000 = (7000)(2. 7)
๐น๐ก − 27000 = 18900
๐น๐ก = 45900๐
c) Find the thrust force needed if the plane must decelerate at a rate of 1.5 m/s2
๐น๐๐๐ก = ๐๐
๐น๐ก − 27000 = (7000)(− 1. 5)
๐น๐ก − 27000 =
− 10500
๐น๐ก = 16500๐
8. A dedicated 88 kg physics student stands on a scale on the elevator
๐๐๐๐๐ ๐๐๐๐๐  ๐๐๐๐๐ ๐๐๐ค๐๐ค๐๐๐๐  ๐ ๐ ๐กโ๐๐ก:
๐๐๐ ๐๐ก๐๐ฃ๐ = ๐ท๐๐ค๐
๐๐๐๐๐ก๐๐ฃ๐ = ๐๐
a) Find the scale reading (in Newtons) when the elevator moves downwards at a steady 2.7 m/s
๐น๐๐๐ก = ๐๐
๐น๐๐๐ก = (88)(9. 8)
๐น๐๐๐ก = 862. 4๐
b) Find the reading when the elevator accelerates downwards at 1.9 m/s2
๐น๐๐๐ก = ๐๐ − ๐๐
๐น๐๐๐ก = (88)(9. 8) − (88)(1. 9)
๐น๐๐๐ก = 862. 4 − 167. 2
๐น๐๐๐ก = 695. 2๐
c) Find the reading when the elevator accelerates upwards at 2.9 m/s2
๐น๐๐๐ก = ๐๐ − ๐๐
๐น๐๐๐ก = (88)(9. 8) − (88)(− 2. 9)
๐น๐๐๐ก = 862. 4 + 255. 2
๐น๐๐๐ก = 1117. 6๐
d) Find the acceleration when the scale reading is 800 N. Which way is the elevator going?
๐น๐๐๐ก = ๐๐ − ๐๐
800 = (88)(9. 8) − 88๐
800 = 862. 4 − 88๐
88๐ = 62. 4
๐ = 0. 71 ๐/๐
๐โ๐ ๐๐๐๐ฃ๐๐ก๐๐ ๐๐  ๐๐๐๐๐๐๐๐๐ก๐๐๐ ๐๐๐ค๐.
e) Find the acceleration when the scale reading is 990 N. Which way is the elevator going?
๐น๐๐๐ก = ๐๐ − ๐๐
990 = (88)(9. 8) − 88๐
990 = 862. 4 − 88๐
88๐ = − 127. 6
๐ = − 1. 45 ๐/๐
๐โ๐ ๐๐๐๐ฃ๐๐ก๐๐ ๐๐  ๐๐๐๐๐๐๐๐๐ก๐๐๐ ๐ข๐.
f) If he uses a scale calibrated in kg, find the reading when the elevator accelerates upwards at 2.9 m/s2
๐น๐๐๐ก = ๐๐ − ๐๐
๐น๐๐๐ก = (88)(9. 8) − (88)(− 2. 9)
๐น๐๐๐ก = 862. 4 + 255. 2
๐น๐๐๐ก = 1117. 6๐
๐ =
๐ =
๐น๐๐๐ก
๐
1117.6
9.8
๐ = 114 ๐๐
9. (challenge) A 500 kg rocket accelerates upward at 3.7 m/s2 near the surface of the earth. Assuming
the thrust force stays constant, what is the upward acceleration when the rocket is exactly one earth
radius above the surface of the earth?
๐น๐๐๐ก = ๐๐
๐น๐๐๐ก = (500)(3. 7)
๐น๐๐๐ก = 1850๐
๐น๐๐๐ก = ๐น๐ก − ๐๐
1850 = ๐น๐ก − (500)(9. 8)
1850 = ๐น๐ก − 4900
๐น๐ก = 6750๐
๐∝
๐∝
๐∝
1
2
๐
1
2
2
1
4
9. 8 &times;
1
4
2
= 2. 45 ๐/๐
๐น๐๐๐ก = ๐๐
๐น๐๐๐ก = ๐น๐ก − ๐๐
๐๐ = ๐น๐ก − ๐๐
500๐ = 6750 − (500)(2. 45)
500๐ = 6750 − 1225
500๐ = 5525
2
๐ = 11. 05 ๐/๐
```