Question

The figure shows a block of mass 1.00 kg is on a frictionless 37.0⁰ incline that is subject
to a horizontal force of 5.00 N. (a) What is its acceleration? (b) If it is initially moving up
the incline at 4.00 m/s, what is its displacement along the incline in 2.10 s?

息The figure shows a block of mass 1.00 kg is on a frictionless 37.0° incline that is subject to a horizontal force of 5.00 N. (a) What is its acceleration? (b) If it is initially moving up the incline at 4.00 m/s, what is its displacement along the incline in 2.10 s? 5 N1 kg △370

Answer 1

Given : M = 1 kg , $\theta$ = 37° , F = 5 N , v_i = 4 m/s , t = 2.10 s

Solution :

(a) Acceleration

According to the given figure , the normal force acting on block by the incline is given by:

R = Mgcos$\theta$

= (1kg)(9.81 m/s²)cos(37°)

= 7.83 N

The force acting in the negative x direction is given by :

F_x' = Mgsin$\theta$

= (1 kg)(9.81 m/s²)sin(37°)

= 5.90 N

The force acting in x direction is given by:

F_x = (5N)cos(37°) = 3.99 N

The net force acting is given by:

F_net = F_x - F'_x

= 3.99 - 5.90

= -1.91 N (negative sign indicates that it acts toward negative x axis)

According to the newton's second law of motion , we have :

F_net = Ma

-1.91N = (1 kg)a

i.e. a = -1.91m/s² [ negative sign indicates that it is sliding down]

Answer : - 1.91 m/s²

(b)

Using equation of motion :

v = v_i + at

v = 4 m/s +(-1.91m/s²)(2.10s)

= -0.011 m/s

Using equation of motion :

d = vt - (1/2)at²

= (-0.011)(2.10) - (1/2)(-1.91)(2.10)²

= 4.19 m

Answer : 4.19 m