Two constant forces act on an object of mass m= 5.4kg moving in the xy plane as

Question

Two constant forces act on an object of mass m= 5.4kg moving in the xy plane as shown in the figure above. Force F1 is 26.5N at 35 degrees, and force F2 is 45.5N at 150 degrees. At time t = 0, the object is at the origin and has velocity (3.30i + 2.00j) m/s.

A.) Express the two forces in unit-vector notation.

B.) Find the total force exerted on the object.

C.) Find the objects acceleration.

**Now, consider the instant t = 3.00s

D.) Find the objects velocity at t= 3s

E.) Find its position at t = 3s

F.) Find its kinetic energy from (0.5)mvf2 at t = 3s

Explanation / Answer

this is very big question i will do initial solution and rest i will guide you how to do the rest

A)

F1=(F1cos35o)i +(F1sin35o)j
find the angle of F2 from the negative X-axis
180-150= 30o
F2=(F2cos30o)i + (F2sin30o)j
here second quadrant X is negative and Y is positive
Just keep it in mind
F2=(-F2cos30o)i + (F2sin30o)j

B)

Take the vector sum
F1=(F1coos35)i +(F1sin35)j
F2=(-F2cos30)i + (F2sin30)j
-----------------------------------
F(total)=(-16.1)i + (40.2)j find the angle in between the resultant
using Tan-1(40.2/-16.1) = -68.2o, your angle is 68.2o
F(total)= SQRT(-16.1)2 + (40.2)2= 40.3 N

SO F(total) = (F(total)cos68.2)i + (F(total)sin68.2)j

c) use Newton's Law F(total) =ma
a= F(total) / mass = 8.4 m/s^2

a= (8.4*cos68.2)i + (8.4*sin68.2)j

d) V= V(initial) * at use trigonometric functions for calculation of V

e) Use X= X(initial) + V(inital)*t + 1/2 at^2   use trigonometric functions for calculation

F) 1/2 m V(final)2
G) ½mvi2 + F •R
where R is the change in distance from the initial position t=0 to the final position where t=3 seconds.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

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