r2 M&I Begin Date: 9/3/2018 12:01:00 AM-Due Date: 9/11/2018 800:00 AM End Date:9

Question

r2 M&I Begin Date: 9/3/2018 12:01:00 AM-Due Date: 9/11/2018 800:00 AM End Date:9728/2018 11:59:00 PM (9%) Problem 1 An object of mas. kg is moving under the effect of a constant force M At initial time , 0 object is known to have the position m and a velocity of m/s with respect to a certain frame of reference. tus us 25% Part (a) what ould the x-comp nent ofthe object's velocity be at time-5 s? V2 D Deductions Potential 100% sind) cos() tan() !(111 7:89 cotano asinacos0 atan) acotan sinh0 cosh0 tanh) cotanh) detailed view O Degrees Radians Igive up 25% Part (b) what would the yoomponent of the object's velocity be at time 1-5 s?

Explanation / Answer

Ans:-

Given data:- m= 45kg, Fx= 5.5N Fy = -2.5N at t=0

X1= -6m, Y1 = 5m, Vix= -12m/s , Viy = 10m/s

Part a] Vfx = ? at t = 5s

Newton’s second law

Fx = m*ax

5.5 = 45*ax

ax=0.12m/s^2

now kinematics equation

Vfx = Vix + ax*t

= -12 + 0.12*5

Vfx= -11.4m/s

Part b] Vfy = ? at t= 5s

Fy = m*ay

-2.5 = 45*ay

ay= -0.056m/s^2

by using kinematics eqn

Vfy = Viy +ay*t

    = 10 + (-0.056*5)

Vfy = 9.72m/s

Part c] X2= ?

Also using kinematics eqn

X2 –X1 = Vix*t + ½ *ax*t^2

X2 –(-6) = -12*5 + ½ *0.12*5^2

X2 + 6 = -60 +1.5

X2= -52.5m

Part d]Y2 =?

Kinematics eqn

Y2-Y1 = Viy*t + ½ ay*t^2

Y2 – 5 = 10*5 + ½ (-0.056)*5^2

Y2-5 = 50 -0.7

Y2=54.3m

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