The relationship between force and position is shown in the given figure (in a one-dimensional case). The work done by the force in displacing a body from \(x = 1~\text{cm}\) to \(x = 5~\text{cm}\) is:
      
1.   \(20~\text{ergs}\) 
2.   \(60~\text{ergs}\)
3.   \(70~\text{ergs}\) 
4.   \(700~\text{ergs}\)

A position dependent force \(F=7-2x+3x^2\) N acts on a small body of mass \(2\) kg and displaces it from \(x = 0\) to \(x = 5\) m. The work done in joule is:

A position-dependent force; \(F=6+8x-3x^2\) N acts on a small body of mass \(3\) kg, displacing it from \(x=0\) to \(x=2\) m. The work done in joule is:
1. \(20\) 
2. \(40\) 
3. \(10\) 
4. \(12\) 

A block of mass \(10\) kg, moving in the \(x\text-\)direction with a constant speed of \(10\) ms^-1, is subjected to a retarding force \(F=0.1x\) J/m during its travel from \(x =20\) m to \(30\) m. Its final kinetic energy will be:

A force \(F = -k(y\hat i +x\hat j)\) (where \(k\) is a positive constant) acts on a particle moving in the \(xy\text-\)plane. Starting from the origin, the particle is taken along the positive \(x\text-\)axis to the point \((a,0)\) and then parallel to the \(y\text-\)axis to the point \((a,a)\). The total work done by the force on the particle is:
1. \(-2ka^2\)
2. \(2ka^2\)
3. \(-ka^2\)
4. \(ka^2\)

The relationship between the force F and the position x of a body is as shown in the figure. The work done in displacing the body from x = 1 m to x = 5 m will be:

The graph between the resistive force \(F\) acting on a body and the distance covered by the body is shown in the figure. The mass of the body is \(25\) kg and the initial velocity is \(2\) m/s. When the distance covered by the body is \(4\) m, its kinetic energy would be:

   
1.   \(50\) J
2. \(40\) J
3. \(20\) J
4.   \(10\) J