For each problem make sure to include a FBD (free body diagram a.k.a. force diag
ID: 1421163 • Letter: F
Question
Explanation / Answer
1) Force(net) = m * a
applied force - friction = x - 90 = 24 *3
x - 90 = 72
x = 162 N = applied force
2) Given:
10 kg = m, mass of the block being pulled
3 m/s^2 = a, acceleration of the block
50 N = Ff, frictional force acting on the block
Find:
Ft = tension on the rope
Solution:
Fnet = ma
Ft - Ff = ma
Ft = ma + Ff
Ft = (10kg)(3 m/s^2) + 50 N
Ft = 30 N + 50N
Ft = 80 N
3) Force = ma = 40 x 5 = 200 N
Frictional force = 210 - 200 = 10 N
5) Mass = 600/9.8 = 61.22 kg
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