Question Video: Calculating the Work Done by a Force in Vector Form Given Its Position-Time Expression | Nagwa Question Video: Calculating the Work Done by a Force in Vector Form Given Its Position-Time Expression | Nagwa

# Question Video: Calculating the Work Done by a Force in Vector Form Given Its Position-Time Expression Mathematics • Third Year of Secondary School

## Join Nagwa Classes

A force 𝐅 = (−4𝐢 − 9𝐣) N is acting on a particle whose position vector as a function of time is given by 𝐫(𝑡) = ((−9𝑡 − 8)𝐢 + (−3𝑡² + 2)𝐣) m. Calculate the work 𝑤 done by the force 𝐅 between 𝑡 = 3 and 𝑡 = 8 seconds.

03:22

### Video Transcript

A force 𝐅 equal to negative four 𝐢 minus nine 𝐣 newtons is acting on a particle whose position vector as a function of time is given by 𝐫 of 𝑡 is equal to negative nine 𝑡 minus eight 𝐢 plus negative three 𝑡 squared plus two 𝐣 meters. Calculate the work 𝑤 done by the force 𝐅 between 𝑡 equals three and 𝑡 equals eight seconds.

In this question, we are asked to find the work done by a force. This is equal to the scalar or dot product of the force vector 𝐅 and displacement vector 𝐝. We are told that the force acting on the particle is equal to negative four 𝐢 minus nine 𝐣 newtons. We are also given the position vector as a function of time and need to calculate the work done between 𝑡 equals three and 𝑡 equals eight seconds. When 𝑡 is equal to three, the position vector 𝐫 of 𝑡 is equal to negative nine multiplied by three minus eight 𝐢 plus negative three multiplied by three squared plus two 𝐣. This simplifies to negative 35𝐢 minus 25𝐣.

When 𝑡 is equal to eight seconds, the position vector of the particle is negative nine multiplied by eight minus eight 𝐢 plus negative three multiplied by eight squared plus two 𝐣. This is equal to negative 80𝐢 minus 190𝐣. We are now in a position to calculate the displacement over this time period. We subtract the position vector at 𝑡 equals three from the position vector at 𝑡 equals eight. Collecting the 𝐢- and 𝐣-components gives us negative 45𝐢 minus 165𝐣.

We now have the force vector in newtons and displacement vector in meters. As already mentioned, the work done is the dot or scalar product of these. And this is equal to the sum of the products of the individual components. We have negative four multiplied by negative 45 plus negative nine multiplied by negative 165. This is equal to 180 plus 1485 which is equal to 1665. As we are working in standard units, the work done by the force 𝐅 between 𝑡 equals three and 𝑡 equals eight seconds is 1665 joules.

## Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

• Interactive Sessions
• Chat & Messaging
• Realistic Exam Questions