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lidiya [134]
3 years ago
14

A 3 kg mass object is pushed 0.6 m into a spring with spring constant 210 N/m on a frictionless horizontal surface. Upon release

, the object moves across the surface until it encounters a rough incline. The object moves UP the incline and stops a height of 1.5 m above the horizontal surface.
(a) How much work must be done to compress the spring initially?
(b) Compute the speed of the mass at the base of the incline.
(c) How much work was done by friction on the incline?
Physics
1 answer:
Svetradugi [14.3K]3 years ago
8 0

Answer with Explanation:

We are given that

Mass of spring,m=3 kg

Distance moved by object,d=0.6 m

Spring constant,k=210N/m

Height,h=1.5 m

a.Work done  to compress the spring initially=\frac{1}{2}kx^2=\frac{1}{2}(210)(0.6)^2=37.8J

b.

By conservation law of energy

Initial energy of spring=Kinetic energy  of object

37.8=\frac{1}{2}(3)v^2

v^2=\frac{37.8\times 2}{3}

v=\sqrt{\frac{37.8\times 2}{3}}

v=5.02 m/s

c.Work done by friction on the incline,w_{friction}=P.E-spring \;energy

W_{friction}=3\times 9.8\times 1.5-37.8=6.3 J

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An infinitely long line of charge has a linear charge density of 7.50×10^−12 C/m . A proton is at distance 14.5 cm from the line
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Answer:

10.22 cm

Explanation:

linear charge density, λ = 7.5 x 10^-12 C/m

distance from line, r = 14.5 cm = 0.145 m

initial speed, u = 3000 m/s

final speed, v = 0 m/s

charge on proton, q = 1.6 x 10^-19 C

mass of proton, m = 1.67 x 10^-27 kg

Let the closest distance of proton is r'.

The potential due t a line charge at a distance r' is given  by

V=-2K\lambda ln\left (\frac{r'}{r}  \right )

where, K = 9 x 10^9 Nm^2/C^2

W = q V

By use of work energy theorem

Work = change in kinetic energy

qV = 0.5m(u^{2}-v^{2})

By substituting the values, we get

V=\frac{mu^{2}}{2q}

-2K\lambda ln\left ( \frac{r'}{r} \right )=\frac{mu^{2}}{2q}

- ln\left ( \frac{r'}{r} \right )=\frac{mu^{2}}{4Kq\lambda }

- ln\left ( \frac{r'}{r} \right )=\frac{1.67 \times 10^{-27}\times 3000\times 3000}{4\times 9\times 10^{9}\times 1.6\times 10^{-19}\times 7.5\times 10^{-12} }

- ln\left ( \frac{r'}{r} \right )=0.35

\frac{r'}{r} =e^{-0.35}

\frac{r'}{r} =0.7047

r' = 14.5 x 0.7047 = 10.22 cm

5 0
3 years ago
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