Home
Class 11
PHYSICS
The work done by a force is given by W =...

The work done by a force is given by W `=vecF.vecs` if `W=0` but `F!=0` and `s!=0` then both `vecF` and `vecs` are

A

in opposite directions

B

in te same direction

C

perpendicular to each other

D

parallel to each other

Text Solution

Verified by Experts

The correct Answer is:
C
`W= vec( F ). vec( s)`
`= (6 hat(i) +2 hat(j) -3 hat(k)) . ( 2 hat(i) - 3 hat(j) + x hat(k))`
But work done = 0
`:. 0 =12 -6 -3x ` `:. 3x = 6 ` `:. X = 2`
Promotional Banner

Topper's Solved these Questions

  • FORCE, WORK AND TORQUE

    MARVEL PUBLICATION|Exercise TEST YOUR GRASP|10 Videos

  • FRICTIONAL IN SOLIDS AND LIQUIDS

    MARVEL PUBLICATION|Exercise TEST YOUR GRASP|10 Videos

Similar Questions

Explore conceptually related problems

The work done by a force is defined by W= vec F cdot vec S cdot It is found that the work done is zero, even if vec F and vec s are not equal to zero. From this we conclude that

The work done by a force vecF during a displacement vecr is given by vecF.vecr . Suppose a force of 12 N acts on a particle in vertically upward directionand the particle is displaced through 2.0 m in vertically downward direction. Find the work done by the force during this displacement.

Work done by a force is given by w=oversetrightarrowF.oversetrightarrowS where oversetrightarrowF is the force and oversetrightarrowS is the displacement. Show that : Work done is also equal to change in K.E.

The linear velocity of a particle on a rotating body is given by vec v = vec omega xx vec r" where "vec omega is the angular velocity and vec r is the radius vector. What is the value of |v| if vec omega = hati -2hatj+2hatk and vec r =4hatj-3hatk ?

A force vecF=hat i+5 hat j+7 hat k acts on a particle and displaces it throught vec s=6 hat I +9 hat k . Calculate the work done if the force is in newton and displacement in metre. (ii) Find the work done by force vec F=2hat i-3 hat j+hatk when its point of application moves from the point A(1,2,-3) to the point B (2,0, -5).

A 0.25 kg puck is initially stationary on an ice surface with negligible friction. At time t = 0, a horizontal force begins to move the puck. The force is given by vecF=(12.0-3.00t^2)hati , with vecF in newtons and t in seconds, and it acts until its magnitude is zero. (a) What is the magnitude of the impulse on the puck from the force between t= 0.750 s and t = 1.25 s? (b) What is the change in momentum of the puck between t=0 and the instant at which F=0?

A force vecF=(3xy-5z)hatj+4zhatk is applied on a particle. The work done by the force when the particle moves from point (0, 0, 0) to point (2, 4, 0) as shown in figure.

Given vecF=(xy^2)hati+(x^2y)hatjN . The work done by vecF when a particle is taken along the semicircular path OAB where the coordinates of B are (4,0) is

If the incident ray on a surface is along the unit vector vec v ,the reflected ray is along the unit vector vec w and the normal is along the unit vector vec a outwards,express vec w in terms of vec a and vec v