l T-Mobile 7:34 PM 35% (10. + A 6.17-kg object passes through the origin at time

Question

l T-Mobile 7:34 PM 35% (10. + A 6.17-kg object passes through the origin at timet0 such that its x component of velocity is 4.70 m/s and its y component of velocity is-3.27 m/s. (a) What is the kinetic energy of the object at this time? (b) At a later time t- 2.00 s, the particle is located at x-8.50 m and y 5.00 m. what constant force acted on the object during this time interval? magnitude (c) What is the speed of the particle at t- 2.00 s? m/s Need Help? ?.Anna A 0.18-kg stone is held 1.4 m above the top edge of a water well and then dropped into it. The well has a depth of 4.8 m. (a) Relative to the configuration with the stone at the top edge of the well, what is the gravitational potential energy of the stone-Earth system before the stone is released? (b) Relative to the configuration with the stone at the top edge of the well, what is the gravitational potential energy of the stone-Earth system when it reaches the bottom of the well? (c) What is the change in gravitational potential energy of the system from release to reaching the bottom of the well? Need Hesp? ?.hnu A 3.00-kg particle moves from the origin to position C, having coordinates x -6.50 m and y 6.50 m (see figure below). One force on the particle is the gravitational force acting in the negative y direction.

Explanation / Answer

1)

mass, m=6.17 kg

vx=4.7 m/sec

vy=-3.27 m/sec

a)

velocity, v=sqrt(vx^2 + vy^2)

=sqrt(4.7^2 + (-3.27)^2)

=5.726 m/sec

kinetic energy, K.E =1/2*m*v^2

=1/2*6.17*5.726^2

=101.148 J

b)

at , t=2 sec,

x=8.5 m and y=5m

use,

x=vx*t+1/2*ax*t^2

8.5=4.7*2+1/2*ax*2^2

==> ax=-0.45 m/sec^2

and

y=vy*t+1/2*ay*t^2

5=(-3.27)*2+1/2*ay*2^2

==> ay=5.77 m/sec^2

the magnitude of acceleration,

a=sqrt(ax^2 + ay^2)

=sqrt(-0.45^2 + 5.77^2)

=5.752 m/sec^2

force acting on the object,

F=m*a

F=6.17*5.752

F=35.49 N

and

tan(theta)=ay/ax

tan(theta)=(5.77/(-0.45))

===>direction, theta=85.54 degrees

c)

after, t=2 sec,

vx=vox+ax*t

vx=4.7+(-0.45)*2

vx=3.8 m/sec^2

and

vy=voy+ay*t

vy=(-3.27)+(5.77)*2

vy=8.27 m/sec^2

==>

v=sqrt(3.8^2 + 8.27^2)

speed, v=9.101 m/sec

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