A12-kg crate rests on a horizontal surface and a boy pulls on itwith a force that is 30? below the horizontal. if the coefficient of static friction is 0.40, theminimum magnitude force he needs to start the crate moving is: a. 44n b. 47n c. 54n d. 56n e. 71n

E.71 N

Explanation:

Given Data

Mass=12 kg

Angle=30°

static friction=0.44 N

To Find

Force=?

Solution

Sum of the forces in the direction of motion is given as

= 0 = F×cos(30) - friction  

Sum of the forces in the vertical direction (up is positive) is given as

= 0 = N - mg - F×sin(30)

N = m×g + F×sin(30)

From the definition of friction

friction = u×N

friction = u×(m×g+F×sin(30))

Plug back into the top

0 = F*cos(30) - (u*m*g+u*F*sin(30))

Solve for F

u×m×g = F×(cos(30) - u×sin(30))

F = (u×m×g) / (cos(30) - u×sin(30))

Put the number we get

F = 71 N

So Option E is correct

d

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)

-4 on