A 20kgbox is being pushed across the floor by a constant force (90,0,0)N. The coefficient of kinetic friction for the table and box is 0.25. At t=5s the box is at location (8,2,-1)m, traveling with velocity (3,0,0)m/s. What is its position and velocity at role="math" localid="1657720192574" t=5.6s?

Short Answer

Expert verified

The position of the box is(x,y,z)=(56.16,2,-1)

The velocity of the box is(14.2,0,0)m/s

Step by step solution

01

Identification of the given data

The mass ofa box is20kg

The constant force is(90,0,0)N

At t = 5,the location of a box is(8,2,-1)m

The travelling velocity is(3,0,0)m/s

Timet=5.6s

02

Determination of its position and velocity

F=ma

Lets assume the components of force(x,y,z)

The force in X direction is 90Nand addition to there is friction also (90-f)

The force in Y direction is0N

The force in Z direction is the force due to gravity force balanced by the normal force ⇒(N-mg=0)

Therefore,

F=ma=(90-f,0,N-mg=0) …(1)

Where,

f=0.25×m×gf=0.25×20×g=5gf=5×10f=50N

Substitute in Equation (1),

F=ma

Where,

F=(90-50,0,0)(40,0,0) which means

a=(2,0,0)

We Know that,

x,y,z=x0,y0,z0+vx0,vy0,vz0t+5at2=8,2,-1+3,0,0t+52,0,0t2=8+3t+t2,2,-1

Put t=5.6sin (x,y,z)

x,y,z=8+3×5.6+5.62,2,-1=56.16,2,-1

Hence, the position of the box is(x,y,z)=(56.16,2,-1)

To find Velocity,

The velocity is the derivative of 8+3t+t2,2,-13+2t,0,0

Put t=5.6,

velocity=(3+2×5.6,0,0)=14.2,0,0m/s

Hence the velocity is(14.2,0,0)m/s

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