Question

The force on a 0.500-kg particle depends on position such that F(r) = (1.00)x2 i +...

The force on a 0.500-kg particle depends on position such that F(r) = (1.00)x2 i + ( 4.00)y j N. If the particle starts from rest at r = (0,0), what will be its speed when it reaches the position r = (4, 3) m?

Please show your work fully and clearly. Thanks!

Homework Answers

Answer #1

F can be resolved into 2 components, Fx along x axis= (1.00)x2 i N and Fy along y axis=  ( 4.00)y j N. Fx moves it along x axis and Fy along y axis. So, work from (0,0) to (4,0) and from (4,0) to (4,3) gives the total amount of work done on the particle . W = Wx + Wy ;where Wx= integral of Fxdx= 1.00*x^3/3 and Wy= integral of Fydy= 4.00*y^2/2

So, W= 1.00/3*4^3 J + 4.00/2*3^2 J
so, W= 39.3 J
Now, 1/2*m*v^2 = 39.3 J { change in kinetic energy is same as the net work done }
so, v^2 = 2/0.500 * 39.3
So, v = 12.5 m/s

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
A particle of mass 2.00 kg moves with position r(t) = x(t) i + y(t) j...
A particle of mass 2.00 kg moves with position r(t) = x(t) i + y(t) j where x(t) = 10t2 and y(t) = -3t + 2, with x and y in meters and t in seconds. (a) Find the momentum of the particle at time t = 1.00 s. (b) Find the angular momentum about the origin at time t = 3.00 s.
PROBLEM 1 The position of a particle on the x axis is given by: x =...
PROBLEM 1 The position of a particle on the x axis is given by: x = 5.00 – 8.00 t + 2.00 t 2 with t in seconds and x in meters. a) Calculate the value of x the moment the particle momentarily stops. b) When t = 0.500 s, is the particle speeding up or slowing down? Explain. PROBLEM 2 A ball is kicked at an angle θ0 with the horizontal with an initial speed of 20.0 m /...
A particle is located at the vector position r with arrow = (9.00î + 12.00ĵ) m...
A particle is located at the vector position r with arrow = (9.00î + 12.00ĵ) m and a force exerted on it is given by F with arrow = (7.00î + 6.00ĵ) N. (a) What is the torque acting on the particle about the origin? τ = N · m (b) Can there be another point about which the torque caused by this force on this particle will be in the opposite direction and half as large in magnitude? Yes...
find the work done by the force field f(x,y)= < x2+y2, -x > on a particle...
find the work done by the force field f(x,y)= < x2+y2, -x > on a particle that moves along the curve c: x2+y2=1, counterclockwise from (0,1) to (-1,0)
The pulley is a uniform cylinder with mass m3 = 0.400 kg and radius R= 4.00...
The pulley is a uniform cylinder with mass m3 = 0.400 kg and radius R= 4.00 cm, the other two masses are m1 = 2.00 kg and m2 = 1.00 kg, and α = 35.0 degrees. Assume the rope is massless, there is no slipping of the rope on the pulley, there is no friction between m1 and the incline, and the incline position is fixed. (a) What is the acceleration of m1 and m2 (both magnitude and direction)? What...
A single conservative force acts on a 5.00 kg particle. The equation Fx = (2x +...
A single conservative force acts on a 5.00 kg particle. The equation Fx = (2x + 4) N describes this force, where x is in meters. As the particle moves along the x axis from x = 3.00 m to x = 7.00 m, calculate the following. (a) the work done by this force on the particle J (b) the change in the potential energy of the system J (c) the kinetic energy the particle has at x = 7.00...
A particle with a mass m = 2.00 kg is moving along the x axis under...
A particle with a mass m = 2.00 kg is moving along the x axis under the influence of the potential energy function U(x) = (2.00 J/m2)x2 − 32.0 J. If the particle is released from rest at the position x = 6.40 m, determine the following. (The sign is important. Be sure not to round intermediate calculations.) (a) total mechanical energy of the particle at any position: =_____ J. (b) potential energy of the particle at the position x...
Force F= (2.84 i - 1.39 k)N acts on a pebble with position vector r= (6.54...
Force F= (2.84 i - 1.39 k)N acts on a pebble with position vector r= (6.54 j - 1.60 k)m , relative to the origin. What is the resulting torque acting on the pebble about (a) the origin and (b) a point with coordinates (4.96 m, 0, -1.43 m)?
A particle moves with position r(t) = x(t) i + y(t) j where x(t) = 10t2...
A particle moves with position r(t) = x(t) i + y(t) j where x(t) = 10t2 and y(t) = -3t + 2, with x and y in meters and t in seconds. (a) Find the average velocity for the time interval from 1.00 s to 3.00 s. (b) Find the instantaneous velocity at t = 1.00 s. (c) Find the average acceleration from 1.00 s to 3.00 s. (d) Find the instantaneous acceleration at t = 1.00 s.
Suppose the position vector for a particle is given as a function of time by r...
Suppose the position vector for a particle is given as a function of time by r (t) = x(t)î + y(t)ĵ, with x(t) = at + b and y(t) = ct2 + d, where a = 1.60 m/s, b = 1.05 m, c = 0.127 m/s2, and d = 1.20 m. (a) Calculate the average velocity during the time interval from t = 2.10 s to t = 4.25 s. (b) Determine the velocity at t = 2.10 s. dDetermine...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT