You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Tension in a String

Three forces ...

Question

Three forces 20√2N,20√2N , 40N are acting along X,Y&Z axes respectively in a 5√2kg mass at rest at the origin . The magnitude of its displacement after 5 sec is

Open in App

Solution

net force on the body = √[(20√2)^2 + (20√2)^2 + (40)^2 ] = √3200 N = 56.56N.

We know that F = m × a

Therefore, a = F/m = 56.56/(5√2) = 7.99 = 8 m/s2.

We know that S= ut + 0.5at^2 .

u =0 (starting from rest).

t=5s (given)

so, S= 0.5 × 8 × 5^2 = 100m.

Suggest Corrections

Similar questions

Q. Three forces

20 \sqrt{2} N, 20 \sqrt{2} N

and

40 N

are acting along

X, Y

and

Z

- axis respectively on a

5 \sqrt{2}

kg mass at rest at origin. The magnitude of its displacement after

5 s

is,

Q. The vertices of a quadrialteral are

A (1, 2, - 1), B (- 4, 2, - 2), C (4, 1, - 5)

and

D (2, - 1, 3)

. Forces of magnitudes

2 N, 3 N, 2 N

are acting at point

A

along the lines

A B, A C

and

A D

respectively. Their resultant is:

Q. The point from where a ball of mass

2 kg

is projected is taken as the origin of the co-ordinate axes. The

x

and

y

components of its displacement are given by

x = 6 t

and

y = 8 t - 5 t^{2}

. Find the average force on it from

t = 0 s

t = 2 s

seconds.

Force $5 \sqrt{2} N$ and $6 \sqrt{2} N$ are acting on a mass of $1000 k g$ at an angle of $60^{\circ}$ to each other. What is the acceleration of the mass after $10$ second from the instant the forces started acting?

Q. A particle of mass

5 kg

is initially at rest. A force starts acting on it along a fixed direction. The magnitude of the force changes with time as shown in the figure. Find the velocity of the particle at the end of

20 s

Join BYJU'S Learning Program

Three forces 20√2N,20√2N , 40N are acting along X,Y&Z axes respectively in a 5√2kg mass at rest at the origin . The magnitude of its displacement after 5 sec is

net force on the body = √[(20√2)^2 + (20√2)^2 + (40)^2 ] = √3200 N = 56.56N. We know that F = m × a Therefore, a = F/m = 56.56/(5√2) = 7.99 = 8 m/s2. We know that S= ut + 0.5at^2 . u =0 (starting from rest). t=5s (given) so, S= 0.5 × 8 × 5^2 = 100m.

net force on the body = √[(20√2)^2 + (20√2)^2 + (40)^2 ] = √3200 N = 56.56N.

We know that F = m × a

Therefore, a = F/m = 56.56/(5√2) = 7.99 = 8 m/s2.

We know that S= ut + 0.5at^2 .

u =0 (starting from rest).

t=5s (given)

so, S= 0.5 × 8 × 5^2 = 100m.