a block of ice with mass 6.20kg is initially at rest on a frictionless, horizont

Question

a block of ice with mass 6.20kg is initially at rest on a frictionless, 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)=(0.193 m/s^2)t^2+(1.92*10^-2m/s^3)t^3

Part a- calculate the velocity of the object at time t=4.30s

part b- calculate the magnitude of F at time t=4.30s

part c- Calcuate the work done by the force F during the first time interval of 4.30s of the motion.

PLease help, need answers! will rate5 stars

Explanation / Answer

v=dx/dt =0.386t+5.76*10^-3 t^2 ;

a) v= 1.7663024 m/s ;

b) F=d(mv)/dt =m*(0.386+11.52*10^-3 * t) = 2.7003232 N ;

c) change in kinetic energy is equal to wor done bu

f ; work=0.5*6.20*1.766^2;

work done = 9.6681436N.m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.