Question

In a hockey game two defencemen simultaneously hit an opposing forward, who has a mass of 80 kg. One defenceman exerts a force on the forward of 475 N, 25° W of N. The other defenceman exerts a force on the forward of 40 N, 60° E of N. Determine the net force acting on the forward and the forward’s acceleration.

Answer 1

here two forces are acting simultaneously

`F_1 = 475 N` 25^0 W of N

`F_2 = 40 N` 60^0 E of N

now we will write the two forces in components form

`F_1 = -475 sin25 \hat i + 475 cos25 \hat j`

`F_1 = -200.7 \hat i + 430.5 \hat j`

`F_2 = 40sin60 \hat i + 40 cos60 \hat j`

`F_2 = 34.6 \hat i + 20 \hat j`

now net force on it will be

`F = F_1 + F_2`

`F = (-200.7 + 34.6) \hat i + (430.5 + 20)\hat j`

`F = -166.1 \hat i + 450.5 \hat j`

so net force is of magnitude

`F = √(166.1^2 + 450.5^2)`

`F = 480.1 N`

direction is given as

`\theta = tan^(-1)(166.1)/(450.5) = 20.2 degree`

so net force is 480.1 N at 20.2 degree W of N

now the acceleration is given by Newton's II law

`F = ma`

`480.1 = 80 a`

`a = 6 m/s^2`