The diagram below shows a block of mass m = 2.00 k g on a frictionless horizonta

Question

The diagram below shows a block of mass m=2.00kg on a frictionless horizontal surface, as seen from above. Three forces of magnitudes F1=4.00N, F2=6.00N, and F3=8.00N are applied to the block, initially at rest on the surface, at angles shown on the diagram. (Figure 1)

The diagram below shows a block of mass m=2.00kg on a frictionless horizontal surface, as seen from above. Three forces of magnitudes F1=4.00N, F2=6.00N, and F3=8.00N are applied to the block, initially at rest on the surface, at angles shown on the diagram. (Figure 1) What is the direction of a? ? In other words, what angle does this vector make with respect to the positive x axis? Express your answer in degrees to two significant figures. How far (in meters) will the mass move in 5.0 s? Express the distance d in meters to two significant figures. What is the magnitude of the velocity vector of the block at t=5.0s? Express your answer in meters per second to two significant figures. In what direction is the mass moving at time t=5.0s? That is, what angle does the velocity vector make with respect to the positive x axis? Express your answer in degrees to two significant figures.

Explanation / Answer

a) in x direction:

Fx = -F3 + F1cos(25)+ F2cps(360-325)

=> Fx = -8 + 4cos(25)+ 6cps(360-325) = 0.540143413882

in y direction:

Fy = F1sin(25)- F2sin(360-325)

=>Fy = 4sin(25)- 6sin(360-325) = -1.750985571142

thus angle = 360 - atan(1.750985571142/0.540143413882) = 287.143929641974 answer(a)

b)a = sqrt(( 0.540143413882* 0.540143413882)+ (1.750985571142*1.750985571142)) = 1.83240426 m/s^2

d = 0.5*a*t^2 = 0.5*1.83240426*5*5 = 22.90505325 m answer(b)

c)v = u + at

=> v = 0 +1.83240426*5 = 9.1620213 m/sec asnwer(c)

d) angle is same = 287.143929641974 degrees

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---