Answer:
Explanation:
a) La fuerza neta que actúa sobre la caja en la dirección vertical es:
Fnet=Fg−f−Fp *sin45 °
aquí Fg representa la fuerza gravitacional, f es la fuerza de fricción, y Fp es la fuerza de empuje.
Fnet=ma
ma=Fg−f−Fp *sin45 °
a=
=0.24 m/s²
Vf =Vi +at
=0.48+0.24*2
Vf=2.98 m/s
b)
Fnet=Fg−f−Fp *sin45 °
=Fg−0.516Fp−Fp *sin45 °
=30-1.273Fp
Fnet=0 (Ya que la velocidad es constante)
Fp=30/1.273
=23.56 N
No one is going to handle that for a mere 5 points lol.
3.258 m/s Explanation: The spring constant is assumed to be 263 N/m and the displacement of the spring is also assumed to be 0.7 m; the coefficient of friction between blocks is 0.4. The energy stored in the spring is described by . Given the conservation of energy in the system, the speed of the 8 kg block just prior to collision is 3.258 m/s.
Answer:
293.7 degrees
Explanation:
A = - 8 sin (37) i + 8 cos (37) j
A + B = -12 j
B = a i + b j, where a and b represent constants to solve for.
A + B = (a - 8 sin (37) ) i + ( 8cos(37) + b ) j
- 12 j = (a - 8 sin (37) ) i + ( 8cos(37) + b ) j
By comparing the coefficients of i and j:
a = 8 sin (37) = 4.81452 m
b = -12 - 8cos(37) = -18.38908
Thus,
B = 4.81452 i - 18.38908 j..... 4th quadrant
<pTherefore,
cos(Q) = 4.81452 / 12
Q = 66.346 degrees
360 - Q gives us 293.65 degrees from the + x-axis in a counterclockwise direction.