A block of ice with mass 6.10kg is initially at rest on a frictionless, horizont
ID: 2190612 • Letter: A
Question
A block of ice with mass 6.10kg is initially at rest on a frictionless, horizontal surface. A worker then applies a horizontal force F to it. As a result, the block moves along the x-axis such that its position as a function of time is given by x(t)=(0.192m/s^2)t^2 + ( 2.05*10^?2 m/s^3)t^3.
a) Calculate the velocity of the object at time t=3.80s
b) Calculate the magnitude of F at time t = 3.80s.
c) Calculate the work done by the force F during the first time interval of 3.80s of the motion.
A block of ice with mass 6.10kg is initially at rest on a frictionless, horizontal surface. A worker then applies a horizontal force F to it. As a result, the block moves along the x-axis such that its position as a function of time is given by x(t)=(0.192m/s^2)t^2 + ( 2.05*10^?2 m/s^3)t^3. a) Calculate the velocity of the object at time t=3.80s b) Calculate the magnitude of F at time t = 3.80s. c) Calculate the work done by the force F during the first time interval of 3.80s of the motion.Explanation / Answer
v=dx/dt =0.384t+6.15*10^-3 t^2 ; a) v= 5.633 m/s ; b) F=d(mv)/dt =m*(0.384+12.30*10^-3 * t) = 2.63 N ; c) change in kinetic energy is equal to wor done bu f ; work=0.5*6.10*5.633^2; work done = 96.78 N.m
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