Question

A force of 6 N is applied and the spring is stretched by 60 mm. What is its elastic force if we stretch the spring by 4 cm?

Construction requires glass with dimensions of 5 x 2 x 0.02 dm and a density of 2500 kg / m3. What is the weight of this glass?

a force of 120 N, 110 N and 300 N is applied to the moving vehicle in one line and in one direction. Determine their representative.

A package of mass 200 kg is pulled with a force of 350 N. How far did he move if 195 J work was done while pulling him?

Answer 1

1) Let's first find the spring constant from the Hooke's Law:

$$k=\dfrac{F}{x}=\dfrac{6\ N}{60\cdot10^{-3}\ m}=100\ \dfrac{N}{m}.$$

Finally, we can find the elastic force if the spring will stretched by 4 cm:

$$F=kx=100\ \dfrac{N}{m}\cdot0.04\ m=4\ N.$$

2)

$$W=mg=\rho Vg,$$ $$W=2500\ \dfrac{kg}{m^3}\cdot0.5\ m\cdot0.2\ m\cdot0.002\ m\cdot9.8\ \dfrac{m}{s^2}=4.9\ N.$$

3) The net force applied to the moving vehicle equals to the sum of the forces:

$$F_{net}=F_1+F_2+F_3,$$ $$F_{net}=120\ N+110\ N+300\ N=530\ N.$$

4) By the definition of the work done, we have:

$$W=Fd,$$ $$d=\dfrac{W}{F}=\dfrac{195\ J}{350\ N}=0.56\ m.$$