Question

# The work done by a force acting on a body is as shown in the graph. what is the total work done in covering an initial distance of 15 m?

- 50 J
- 75 J
- 100 J
- 25 J

## The correct answer is: 75 J

### Related Questions to study

### A 8 kg mass moves along x - axis. Its accelerations as a function of its position is shown in the figure. What is the total work done on the mass by the force as the mass moves from x=0 to x=6 cm ?

### A 8 kg mass moves along x - axis. Its accelerations as a function of its position is shown in the figure. What is the total work done on the mass by the force as the mass moves from x=0 to x=6 cm ?

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We have to be careful while solving quadratic equation questions as there are chances of mistakes with the signs while finding the roots. Sometimes the students try to use factorisation methods to solve the quadratic equation, but in this type of questions, it is not recommended at all. Always try to use the quadratic formula for solving. You can also remember the statement to be proved, given in the question as a property for two positive numbers.

### If A and G are A.M and G.M between two positive numbers a and b are connected by the relation A+G=a-b then the numbers are in the ratio

We have to be careful while solving quadratic equation questions as there are chances of mistakes with the signs while finding the roots. Sometimes the students try to use factorisation methods to solve the quadratic equation, but in this type of questions, it is not recommended at all. Always try to use the quadratic formula for solving. You can also remember the statement to be proved, given in the question as a property for two positive numbers.

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### The relationship between force and position is shown in the figure given (in one dimensional case) calculate the work done by the force in displacing a body from x=0 cm to x=5 cm

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Here note that the fifth term is having fourth power of 2 and not fifth power. We need not to find all nine terms separately; only finding the product is enough because that product will then be written in the form of the term that is known. Terms in a G.P. are having a common ratio in between. That’s why the power of r is increasing as the terms are increasing.

### Fifth term of G.P is 2 The product of its first nine terms is

Here note that the fifth term is having fourth power of 2 and not fifth power. We need not to find all nine terms separately; only finding the product is enough because that product will then be written in the form of the term that is known. Terms in a G.P. are having a common ratio in between. That’s why the power of r is increasing as the terms are increasing.