A crate weighing 523 N rests on a plank that makes a 22.0  angle with the ground. Find the components of the crate's weight force parallel and perpendicular to the plank. ( Fgx = –196 N, Fgy = –485 N)

Explanation:

The weight of the crate will e equal to its normal reaction R

Since R = mgsin theta = W

W = mgsintheta ( resolved along the vertical perpendicular to the plank

Fy = 523sin22°

Fy = 195.92N

Since the weight is acting along the negative y direction,

Fy = -195.92N

The force parallel to the plank will be the force acting in the horizontal direction.

Fx = Fcos theta

Fx = 523cos 22°

Fx = 484.9N

no

sodium is solid at rt

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s