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A single constant force $\mathbf{F}=(3 \mathbf{i}+5 \mathbf{j}) \mathrm{N}$ acts on a $4.00-\mathrm{kg}$ particle. (a) Calculate the work done by this force if the particle moves from the origin to the point having the vector position $\mathbf{r}=(2 \mathbf{i}-3 \hat{\mathbf{j}}) \mathrm{m} .$ Does this result depend on the path? Explain. (b) What is the speed of the particle at r if its speed at the origin is 4.00 $\mathrm{m} / \mathrm{s}$ ? (c) What is the change in the potential energy?

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What done it was f dot delta are equals three i plus five j newton dot to I minus three j major. So that comes with three times two equals six, five times three equals minus 15 Jews. It was negative. Nine shoes now walk down. It was increased in kinetic energy. So we have increase in kinetic energy half times. Mass equals four times of the square equals Does the final kind of big energy on the initial kinetic energies have times four times initial velocity squared equals negative nine. So solving days we find V equals 3.4 middle per second and increasing potential energy is negative off the work done equals 9 June.

Indian Institute of Technology Kharagpur