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When an object is pushed by a force and moves in the direction of the force, work is done on the object. The amount of work equals the force times the distance ([tex]W = Fd[/tex]).

A small particle is pushed by a force of [tex]2.6 \times 10^4[/tex] newtons over a distance of [tex]3.7 \times 10^3[/tex] meters. How much work was done on the particle? Express the answer to the correct number of significant figures and in proper scientific notation.

The force did [tex] \times 10^ [/tex] newton-meters of work on the particle.

Sagot :

To solve this problem, let's follow the given formula for work done:

[tex]\[ W = F \times d \][/tex]

Where:
- [tex]\( W \)[/tex] is the work done,
- [tex]\( F \)[/tex] is the force applied, and
- [tex]\( d \)[/tex] is the distance over which the force is applied.

Given:
- The force applied [tex]\( F = 2.6 \times 10^4 \)[/tex] newtons,
- The distance [tex]\( d = 3.7 \times 10^3 \)[/tex] meters.

Step-by-step solution:

1. Multiply the Force and Distance:

[tex]\[ W = (2.6 \times 10^4 \, \text{N}) \times (3.7 \times 10^3 \, \text{m}) \][/tex]

2. Carry out the multiplication:

[tex]\[ W = 2.6 \times 3.7 \times 10^4 \times 10^3 \][/tex]

3. Multiply the coefficients (2.6 and 3.7):

[tex]\[ 2.6 \times 3.7 = 9.62 \][/tex]

4. Add the exponents of 10 (because when you multiply powers of 10, you add the exponents):

[tex]\[ 10^4 \times 10^3 = 10^{4+3} = 10^7 \][/tex]

So,

[tex]\[ W = 9.62 \times 10^7 \, \text{newton-meters (or joules)} \][/tex]

Therefore, the work done on the particle is [tex]\( 9.62 \times 10^7 \)[/tex] newton-meters.