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
𝐹 = 𝑚𝑎
220 = 70𝑎
2
𝑎 = 3. 1 𝑚/𝑠
b) find the velocity after 10 seconds.
𝑣 = 𝑣0 + 𝑎𝑡
𝑣 = 0 + (3. 1)(10)
𝑣 = 31 𝑚/𝑠
2. A 561kg rocket has a thrust force of 18500N. It starts from rest on the ground. Find its height after 10
seconds.
𝐹 = 𝑚𝑎 − 𝑚𝑔
18500 = 561𝑎 − 561(9. 8)
2
𝑎 = 33. 0 𝑚/𝑠
𝑎𝑛𝑒𝑡 = 33. 0 − 9. 8
2
𝑎𝑛𝑒𝑡 = 23. 2 𝑚/𝑠
ℎ = 𝑣0𝑡 +
1
2
2
𝑎𝑡
ℎ = (10)(10) +
ℎ = 0 + 1160
ℎ = 1160 𝑚
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
𝐹𝑛𝑒𝑡 = 2700𝑁 − 750𝑁 − 300𝑁
𝐹𝑛𝑒𝑡 = 2650𝑁
b) Find the acceleration.
𝐹𝑛𝑒𝑡 = 𝑚𝑎
2650 = 530𝑎
2
𝑎 = 5 𝑚/𝑠
4. A 40kg sled is pulled with a force of 250N and accelerates at 1.5m/s2.
a) Find the force of friction.
𝐹𝑛𝑒𝑡 = 𝑚𝑎
250 − 𝑓 = (40)(1. 5)
250 − 60 = 𝑓
𝑓 = 190𝑁
b) Find the coefficient of friction
𝑁 = 𝑚𝑔
𝑁 = (40)(9. 8)
𝑁 = 392𝑁
𝑓
μ = 𝑁
μ =
190
392
μ = 0. 48
c) Find the acceleration of the sled if the person pulling lets go
𝐹𝑛𝑒𝑡 = 𝑚𝑎
− 190 = 40𝑎
𝑎 =
2
− 4. 75 𝑚/𝑠
5. a) Find the net force needed to accelerate a 1200 kg car from 20 to 30 km/h in 2.7 seconds
20 𝑘𝑚/ℎ ×
30 𝑘𝑚/ℎ ×
1000 𝑚
1 𝑘𝑚
1000 𝑚
1 𝑘𝑚
×
×
1ℎ
3600 𝑠
1ℎ
3600 𝑠
= 5. 67 𝑚/𝑠
= 8. 33 𝑚/𝑠
𝑣 = 𝑣0 + 𝑎𝑡
8. 33 = 5. 67 + 2. 7𝑎
2. 77 = 2. 7𝑎
2
𝑎 = 1. 029 𝑚/𝑠
𝐹𝑛𝑒𝑡 = 𝑚𝑎
𝐹𝑛𝑒𝑡 = (1200)(1. 029)
𝐹𝑛𝑒𝑡 = 1235𝑁
b) What is the minimum coefficient of friction which allows this acceleration?
𝑁 = 𝑚𝑔
𝑁 = (1200)(9. 8)
𝑁 = 11760𝑁
𝑓 = 𝐹𝑛𝑒𝑡
𝑓 = 1235𝑁
𝑓
μ = 𝑁
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.
𝐹𝑛𝑒𝑡 = 𝑚𝑎
120 = 15𝑎
2
𝑎 = 8 𝑚/𝑠
b) There is a 50N friction force
𝐹𝑛𝑒𝑡 = 𝑚𝑎
120 − 50 = 15𝑎
70 = 15𝑎
2
𝑎 = 4. 7 𝑚/𝑠
c) μ =0.3
𝑁 = 𝑚𝑔
𝑁 = (15)(9. 8)
𝑁 = 147𝑁
𝑓 = μ𝑁
𝑓 = (0. 3)(147)
𝑓 = 44. 1𝑁
𝐹𝑛𝑒𝑡 = 𝑚𝑎
120 − 44. 1 = 15𝑎
75. 9 = 15𝑎
2
𝑎 = 5. 06 𝑚/𝑠
7. A 7000kg jet accelerates forwards against a 27 000N drag force.
a) Find the acceleration if the thrust is 39000 N
𝐹𝑛𝑒𝑡 = 𝑚𝑎
39000 − 27000 = 7000𝑎
12000 = 7000𝑎
2
𝑎 = 1. 7 𝑚/𝑠
b) Find the Thrust required to produce an acceleration of 2.7 m/s2
𝐹𝑛𝑒𝑡 = 𝑚𝑎
𝐹𝑡 − 27000 = (7000)(2. 7)
𝐹𝑡 − 27000 = 18900
𝐹𝑡 = 45900𝑁
c) Find the thrust force needed if the plane must decelerate at a rate of 1.5 m/s2
𝐹𝑛𝑒𝑡 = 𝑚𝑎
𝐹𝑡 − 27000 = (7000)(− 1. 5)
𝐹𝑡 − 27000 =
− 10500
𝐹𝑡 = 16500𝑁
8. A dedicated 88 kg physics student stands on a scale on the elevator
𝑆𝑐𝑎𝑙𝑒 𝑟𝑒𝑎𝑑𝑠 𝑓𝑜𝑟𝑐𝑒 𝑑𝑜𝑤𝑛𝑤𝑎𝑟𝑑𝑠 𝑠𝑜 𝑡ℎ𝑎𝑡:
𝑃𝑜𝑠𝑖𝑡𝑖𝑣𝑒 = 𝐷𝑜𝑤𝑛
𝑁𝑒𝑔𝑎𝑡𝑖𝑣𝑒 = 𝑈𝑝
a) Find the scale reading (in Newtons) when the elevator moves downwards at a steady 2.7 m/s
𝐹𝑛𝑒𝑡 = 𝑚𝑔
𝐹𝑛𝑒𝑡 = (88)(9. 8)
𝐹𝑛𝑒𝑡 = 862. 4𝑁
b) Find the reading when the elevator accelerates downwards at 1.9 m/s2
𝐹𝑛𝑒𝑡 = 𝑚𝑔 − 𝑚𝑎
𝐹𝑛𝑒𝑡 = (88)(9. 8) − (88)(1. 9)
𝐹𝑛𝑒𝑡 = 862. 4 − 167. 2
𝐹𝑛𝑒𝑡 = 695. 2𝑁
c) Find the reading when the elevator accelerates upwards at 2.9 m/s2
𝐹𝑛𝑒𝑡 = 𝑚𝑔 − 𝑚𝑎
𝐹𝑛𝑒𝑡 = (88)(9. 8) − (88)(− 2. 9)
𝐹𝑛𝑒𝑡 = 862. 4 + 255. 2
𝐹𝑛𝑒𝑡 = 1117. 6𝑁
d) Find the acceleration when the scale reading is 800 N. Which way is the elevator going?
𝐹𝑛𝑒𝑡 = 𝑚𝑔 − 𝑚𝑎
800 = (88)(9. 8) − 88𝑎
800 = 862. 4 − 88𝑎
88𝑎 = 62. 4
𝑎 = 0. 71 𝑚/𝑠
𝑇ℎ𝑒 𝑒𝑙𝑒𝑣𝑎𝑡𝑜𝑟 𝑖𝑠 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑛𝑔 𝑑𝑜𝑤𝑛.
e) Find the acceleration when the scale reading is 990 N. Which way is the elevator going?
𝐹𝑛𝑒𝑡 = 𝑚𝑔 − 𝑚𝑎
990 = (88)(9. 8) − 88𝑎
990 = 862. 4 − 88𝑎
88𝑎 = − 127. 6
𝑎 = − 1. 45 𝑚/𝑠
𝑇ℎ𝑒 𝑒𝑙𝑒𝑣𝑎𝑡𝑜𝑟 𝑖𝑠 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑛𝑔 𝑢𝑝.
f) If he uses a scale calibrated in kg, find the reading when the elevator accelerates upwards at 2.9 m/s2
𝐹𝑛𝑒𝑡 = 𝑚𝑔 − 𝑚𝑎
𝐹𝑛𝑒𝑡 = (88)(9. 8) − (88)(− 2. 9)
𝐹𝑛𝑒𝑡 = 862. 4 + 255. 2
𝐹𝑛𝑒𝑡 = 1117. 6𝑁
𝑚 =
𝑚 =
𝐹𝑛𝑒𝑡
𝑔
1117.6
9.8
𝑚 = 114 𝑘𝑔
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?
𝐹𝑛𝑒𝑡 = 𝑚𝑎
𝐹𝑛𝑒𝑡 = (500)(3. 7)
𝐹𝑛𝑒𝑡 = 1850𝑁
𝐹𝑛𝑒𝑡 = 𝐹𝑡 − 𝑚𝑔
1850 = 𝐹𝑡 − (500)(9. 8)
1850 = 𝐹𝑡 − 4900
𝐹𝑡 = 6750𝑁
𝑔∝
𝑔∝
𝑔∝
1
2
𝑟
1
2
2
1
4
9. 8 ×
1
4
2
= 2. 45 𝑚/𝑠
𝐹𝑛𝑒𝑡 = 𝑚𝑎
𝐹𝑛𝑒𝑡 = 𝐹𝑡 − 𝑚𝑔
𝑚𝑎 = 𝐹𝑡 − 𝑚𝑔
500𝑎 = 6750 − (500)(2. 45)
500𝑎 = 6750 − 1225
500𝑎 = 5525
2
𝑎 = 11. 05 𝑚/𝑠