A 5.67-kg object passes through the origin at time t = 0 such that its x compone

Question

A 5.67-kg object passes through the origin at time t = 0 such that its x component of velocity is 5.15 m/s and its y component of velocity is -3.18 m/s. (a) What is the kinetic energy of the object at this time? 80.23 Your response differs from the correct answer by more than 10%. Double check your calculations. J (b) At a later time t = 2.00 s, the particle is located at x = 8.50 m and y = 5.00 m. What constant force acted on the object during this time interval? magnitude direction º measured from the +x axis N (c) What is the speed of the particle at t = 2.00 s? m/s Need Help? Read It

Explanation / Answer

part A :

Vx = 5.15 m/s

Vy = -3.18 m/s

V^2 = Vx^2 + Vy^2

V^2 = 5.15^2 + (-3.18^2)

V^2 = 16.41 (m/s)^2

KE = 0.5 mV^2

KE = 0.5 * 5.67* 16.41

KE = 46.52 Joules

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part B:

use Sx = Uxt + 0.5 ax t^2

at t= 2 s, Sx = 8.5 m

so 8.5 = 5.15 *2 + 0.5 * ax * 2^2

ax = -0.9 m/s^2

Sy = Uyt + 0.5 ay t^2

5 = (-3.18*2) + (0.5* ay * 4)

ay = 5.68 m/s^2

net accleration a^2 = ax^2 + ay^2

a^2 = (-0.9^2) + (5.68)^2

a = 5.75 m/s^2

net force F = ma

F = 5.67 * 5.75 = 32.6N

direction tan theta= ay/ax

tantheta = 5.75/-0.9 = -6.388

theta = 98.89 deg

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part C:

Vx = Vox + ax t

Vx = 5.15 + (-0.9* 2) = 3.35 m/s

Vy = Voy + ayt

Vy = (-3.18) + (5.68 * 2)

Vy = 8.18 m/s

Vnet^2 = Vx^2 + Vy^2

Vnet^2 = 3.35^2 + 8.18^2

Vnet= 8.84 m/s

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