Question

work =Force*displacemment.if force =10N, and costeta=60degree then then find the work =force*d*cos60.dont find d(displacement)

Answer 1

The second case is easy. Work is defined as the dot product between force and displacement. It is given that a constant100 N force was applied until the object moved 10 m away, the work done would be simply:
W = F*d = 100*10 = 1000 J 
(Assuming displacement is in direction of the force at all times)

The first case is a little more tricky.
From what I understand, the first case essentially means that an instantaneous force of 100 N was applied at time t=0and then the body was left to move. It was like when a body is just given a little kick and then left alone to move or how a ball is hit by a bat and then the ball moves itself up until a certain distance. 

For mathematical treatment, I will call the force in this case as F=100δ(t)F=100δ(t)
( δ(t)δ(t) is the familiar Dirac delta function which is zero for all values of t other than when t=0. At t=0, δ(t)=δ(0)=1δ(t)=δ(0)=1 )
So our force becomes 100 N only at timet=0 and is zero afterwards.

Now the force though defined to be instantaneous is not actually instantaneous and acts for a very small time. By invoking the Dirac delta function, we can say that the force acted for an infinitesimally small amount of time.
There is a quantity called impulse which is the force times the time during which it acts. This impulse is also equal to the change in momentum.

Fdt=mdvFdt=mdv
Integrating both sides and considering that the impact acted for a very small time t0t0 resulting in a velocity v at the end.
∫t00100δ(t)dt=m∫v0dv∫0t0100δ(t)dt=m∫0vdv
The left hand side integral is zero only except at time t=0.
Therefore, it becomes:
100=mv100=mv which implies that v = 100/m   m/s

The object got a kick resulting in an initial velocity equal to 100/m m/s. Due to friction and air resistance, it got slowed down and eventually stopped moving after having moved 10 m. 

The work done would be the change in kinetic energy as the object moved 10 m.
Work done = change in KE = 1/2∗m(10000/m2−0)=5000/m1/2∗m(10000/m2−0)=5000/mJoules.

The work done in this case will depend on the mass of the object.