$K.E = \frac{1}{2} m {v}^{2} \\ {v}^{2}_i = {v}^{2} _x + {v}^{2} _y \\ = {(6)}^{2} + {( - 2)}^{2} = 36 + 4 = 40m. {s}^{ - 1} \\ K.E_i = \frac{1}{2} (3) (40) = 60J \\ \\ {v}^{2}_f = {v}^{2} _x + {v}^{2} _y \\ = {(8)}^{2} + {(4)}^{2} = 80m. {s}^{ - 1} \\ K.E_i = \frac{1}{2} (3) (80) = 120J \\ W_{net}=K.E_f-K.E_i \\ = 120J - 60J \\ = 60J$

3 0

3 years ago

A certain spring stretches 3 cm when a load of 15 n is suspended from it. how much will the spring stretch if 30 n is suspended

Alik [6]

Initially, the spring stretches by 3 cm under a force of 15 N. From these data, we can find the value of the spring constant, given by Hook's law:
$k= \frac{F}{\Delta x}$
where F is the force applied, and $\Delta x$ is the stretch of the spring with respect to its equilibrium position. Using the data, we find
$k= \frac{15 N}{3.0 cm}=5.0 N/cm$

Now a force of 30 N is applied to the same spring, with constant k=5.0 N/cm. Using again Hook's law, we can find the new stretch of the spring:
$\Delta x = \frac{F}{k}= \frac{30 N}{5.0 N/cm}=6 cm$

4 0

3 years ago

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