A4.00-kg block starts from rest at the top of a 30.0° incline and slides a distance of 2.10 m down the incline in 1.60 s. (a) find the magnitude of the acceleration of the block. m/s2 (b) find the coefficient of kinetic friction between block and plane. (c) find the friction force acting on the block. magnitude n direction (d) find the speed of the block after it has slid 2.10 m.

a) a=1.64 m/s² : magnitude of the acceleration of the block

b) μk = 0.384 : coefficient of kinetic friction

c) Ff = 13.04 N  : in direction (-x), forming a 30° angle with the horizontal.

d) v= 2.62 m/s  : speed of the block after it has slid 2.10 m.

Explanation:

kinematic analysis

Because the block moves with uniformly accelerated movement we apply the following formulas:

d= v₀t+ (1/2)*a*t² Formula (1)

vf²=v₀²+2*a*d   Formula (2)

Where:  

d:displacement in meters (m)  

t : time in seconds (s)

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s²

a) v₀=0 ,d= 2.1 m , t= 1.6 s , a=?

We apply the formula (1)

d= v₀t+ (1/2)*a*t²

2.1= 0 +  (1/2)*a*(1.6)²

2.1= 1.28*a

a= (2.1)/(1.28)

a= 1,64 m/s²

b)Kinetics analysis

We apply Newton's second law:

∑F = m*a (Formula 3)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Data:

m=4 kg

β = 30°

g = 9.8m/s²

W= m*g = 4 kg*9.8m/s²= 39.2 N : Block weight

x-y weight components

Wx= Wsinβ=39.2 N*sin30° = 19.6 N

Wy= Wcosβ=39.2 N*cos30° =33.95 N

Look at the attached graphic

We apply the formula (3) to calculate the frictional force (Ff)

∑Fx = m*a

Wx-Ff=m*a

Wx-m*a= Ff

Ff = 19.6 - 4* 1.64

Ff =13.04 N

We apply the formula (3) to calculate theNormal force (N)

∑Fy = m*a,  a =0 in direction y

N-Wy = 0

N = Wy = 33.95 N

b)Friction force formula

Ff =μk* N

μk = (Ff) / (N)

μk = (13.04) / (33.95)

μk = 0.384

c) Ff =μk* N

  Ff = 0.384*33.95= 13.04 N  : in direction (-x), forming a 30° angle with the horizontal.

d) Calculation of the final speed at 2.1 m

We apply formula (2) :

vf²=v₀²+2*a*d  

note: the given question is incomplete.

complete question is given below.

which is an example of kinetic energy?

a. the energy stored in ethanol  

b. a hockey puck sliding across the ice  

c. a compressed spring  

d. a ball sitting at the top of a ramp

solution: the kinetic energy is the energy posses by the virtue of the motion of the object.

in the given options, a, c and d are the example of potential energy because it is the energy posses by the virtue of the rest of the object. but option b has the kinetic energy as the object is in motion.

hence, a hockey puck sliding across the ice   is the apex example of kinetic energy.

so, the option (b) a hockey puck sliding across the ice   is the correct option.