Question

A body of mass 2kg is slowly moved up a track in the vertical plane from A to B by a force vec F which at each point is directed along the tangent to the trajectory.The coefficient of friction is 0.8 W_(F): work done by force bar(F) W_(g) : work done by gravity and W_(fr): work done by friction on the block then (g=10m/s^(2))​

Answer 1

Explanation:

Since the acceleration

a is constant, the velocity v

and the position x of the rocket

are

v = at + v(0) ,

x = at2/2 + v(0)t + x(0) .

x

ϕ

Applying the initial conditions yields

v(0) = 0 → v = at,

x(0) = 0 → x =

at2

2

.

The angle ϕ of the radar screen follows as

tan ϕ =

x

l

→ ϕ(t) = arctan

at2

2l

.

Differentiation leads to the angular velocity

ϕ˙(t) = at

l

/

"

1 +

at2

2l

2

#

and the angular acceleration

ϕ¨(t) =

a

l

−

3a

3

t

4

4l

3

/

"

1 +

at2

2l