A box of mass 3.1kg slides down a rough vertical wall. The gravitational force on the box is 30N . When the box reaches a speed
of 2.5m/s , you start pushing on one edge of the box at a 45° angle (use degrees in your calculations throughout this problem) with a constant force of magnitude Fp = 23N . There is now a frictional force between the box and the wall of magnitude 13N . How fast is the box sliding 2.0s after you started pushing on it?
Assuming that the angle at which you push on the edge of the box is again 45∘, with what magnitude of force Fp should you push if the box were to slide down the wall at a constant velocity? Note that, in general, the magnitude of the friction force will change if you change the magnitude of the pushing force. Thus, for this part, assume that the magnitude of the friction force is f=0.516Fp.
Assuming that the angle at which you push on the edge of the box is again 45∘, with what magnitude of force Fp should you push if the box were to slide down the wall at a constant velocity? Note that, in general, the magnitude of the friction force will change if you change the magnitude of the pushing force. Thus, for this part, assume that the magnitude of the friction force is f=0.516Fp.
2 answers:
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
Answer:
(después de dos segundos)
(sin aceleración)
Explanation:
Datos
(constante)
(fricción)
Primero, debemos calcular la velocidad después de 2 segundos, teniendo en cuenta que la velocidad inicial es . Se puede determinar la aceleración usando la Segunda Ley de Newton
Fuerzas verticales:
Donde , es deducido del ángulo recto formado (consultar la imagen anexa).
Luego,
Ahora, resolvemos para
Luego, utilizamos la fórmula para calcular la velocidad después de 2 segundos
Ahora, si la fuerza de fricción es y la caja desciende manteniendo velocidad constante, ¿cuál sería la magnitud de
para conseguirlo?
En este caso, la suma de las fuerzas verticales sería
Donde
Observa que la fuerza neta vertical es cero, ya que no hay aceleración, la caja se mueve a velocidad constante.
Ahora resolvemos para
Por lo tanto, la fuerza necesaria para que no haya aceleración es de 24.5 N.
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