Question

2. A 2kg puck is located at position (5???? 2???)m with velocity (?7???? ???)m/s at time t=0s. I apply a
constant force of magnitude 10N in the direction (?4???+ 2???) for 10 seconds. At time t=10s, the
puck is at position (????+ 10???)m with a velocity of 5m/s in the direction (???+ ???).
a. How much work did I do on the puck?
b. How much work is done on the puck total?
c. What is the average power input to the puck from all sources?

Answer 1

Force = 10 * ( - 4i + 2j) / sqrt (4^2 + 2^2) = -8.94427 i + 4.4721359 j
acceleration = Force/mass = (   -8.94427 i + 4.4721359 j )/2 = -4.472135 i + 2.236067 j
so, displacement = ((-7 *10) + (0.5 * -4.472135 * (10^2)) i + ((-1*10) + (0.5 *2.236067*(10^2)) j = -293.60675 i + 101.80335 j
a) work did I do on the puck = (-8.94427 i + 4.4721359 j ) . (-293.60675 i + 101.80335) = 3081.37646 J
b) we have
displacement = ((vx *t) + (0.5 *ax * (t^2)) i + ((vy*t) + (0.5 *ay*(t^2)) j
so, vx = -7 ; vy = -1 ; t = 10 ;
displacement = (-i + 10 j) = ((-7 *10) + (0.5 *ax * (10^2)) i + ((-1*10) + (0.5 *ay*(10^2)) j
ax = 1.38 m/s^2 ; ay = 0.4 m/s^2
total acceleration = 1.38 i + 0.4 j
total force = mass * acceleration = 2.76i + 0.8 j
work is done on the puck total = total force. displacement
work is done on the puck total = (2.76i + 0.8 j ). (-i + 10 j) = 5.24 J
c)
average power input to the puck from all sources = change in momentum = mass * ( final velocity - initial velocity ) = 2 * (( (5/1.414) +7) i + ((5/1.414)+1) j) = 21.072 i + 9.07213 j
magnitude = sqrt (21.072^2 + 9.07213^2) = 22.94194 W