Download SOLUCIÓN PROBLEMAS DE FUERZAS F= 20 N M= 5 gr= 0.005kg t

SOLUCIÓN PROBLEMAS DE FUERZAS
1. F= 20 N
𝐹
M= 5 gr= 0.005kg
20𝑁
𝑎 = 𝑚 = 0.005 𝑘𝑔 = 40000𝑚/𝑠 2
2. M= 4 Kg
𝑎=
𝐹
𝑚
=
10𝑁
4 𝑘𝑔
F=10N
𝑥=
vi=0
−𝐹𝑟 + 𝐹 = 𝑚. 𝑎
4. F= ?
𝑎𝑡 2
2
t=5seg
= 2.5𝑚/𝑠 2
3. F= 10 N
t =10s
=
40000(102 )
2
x=?
= 200000𝑚 = 200𝐾𝑚
a=?
vf=?
𝑣𝑓 = 𝑎𝑡 = 2.5𝑚/𝑠 2 (5𝑠𝑒𝑔) = 12.5𝑚/𝑠
m=400g
𝑡=
𝑣𝑓
𝑎
m=300Kg
t =?
vf=20m/s
20𝑚/𝑠
= 25𝑚/𝑠2 = 0.8𝑚/𝑠 2
vf =36km/h=10m/s
t=5 seg
𝑣𝑓 10𝑚/𝑠
=
= 2𝑚/𝑠 2
𝑡
5𝑠𝑒𝑔
−𝐹𝑟 + 𝐹 = 𝑚. 𝑎
𝐹 = 𝑚. 𝑎 + 𝐹𝑟
𝑎=
𝐹 = 300𝑘𝑔(2𝑚/𝑠 2 ) + 80𝑁 = 600𝑁 + 80 𝑁 = 680𝑁
5. F= 120 N
𝐹
m=25Kg
120𝑁
𝑎 = 𝑚 = 25 𝑘𝑔 = 4.8𝑚/𝑠 2
6. F= 200N
𝑡=
m=10Kg
t =?
𝑣𝑓
𝑎
vf=90km/h= 25m/s
25𝑚/𝑠
= 4.8𝑚/𝑠2 = 5.2𝑚/𝑠 2
a=?
Fr= 10N
−𝐹𝑟 + 𝐹 = 𝐹 𝑡𝑜𝑡𝑎𝑙
−10𝑁 + 200𝑁 = 190𝑁
7. F= 300 N
m=100Kg
𝐹 = 𝑚. 𝑎
𝐹 = 100𝑘𝑔(2𝑚/𝑠 2 )
𝐹 = 200𝑁
𝐹𝑡𝑜𝑡𝑎𝑙 = 200𝑁 + 300𝑁 = 500𝑁
𝑎=
𝐹
𝑀
=
190𝑁
410𝑘𝑔
Ftotal=?
= 19/𝑠 2
a=2m/s2
Fr= 80N
8. F= 100N
𝐹
m=5Kg vf =?
t=5 seg
90𝑁
𝑎 = 𝑚 = 5 𝑘𝑔 = 18𝑚/𝑠 2
𝑥=
Fr= 10N
x=?
𝑣𝑓 = 𝑎𝑡 = 218𝑚/𝑠 2 (5𝑠𝑒𝑔) = 90𝑚/𝑠
𝑎𝑡 2
18𝑚/𝑠 2 (5𝑠𝑒𝑔2 )
=
= 225𝑚
2
2
9. F= ?
vi=144km/h = 40m/s
a=?
m=200Kg
x=20m
vf=0m/s
2𝑎𝑥 = 𝑣𝑓 2 − 𝑣𝑖 2
𝑎=
𝑣𝑓 2 − 𝑣𝑖 2
0 − (40𝑚/𝑠)2 −1600𝑚2 /𝑠 2
=
=
= −40𝑚/𝑠 2
2𝑥
40𝑚
40𝑚
𝐹 = 𝑚. 𝑎 = 200𝐾𝑔(−40𝑚/𝑠 2 ) = −8000𝑁
10. F= 4500N
Fr=1300N
a=?
m=860Kg
t=12 seg
vf=?
𝐹𝑡𝑜𝑡𝑎𝑙 = 4500𝑁 − 1300𝑁 = 3200𝑁
𝐹
3200𝑁
𝑎 = 𝑚 = 860 𝑘𝑔 = 3.72𝑚/𝑠 2
3.72𝑚
𝑣𝑓 = 𝑎𝑡 = 𝑠2 (12𝑠𝑒𝑔) = 44.65𝑚/𝑠 = 160.7𝐾𝑚/ℎ
11. m=600 Kg
1er tramo 0 – 30 segundos
Como la velocidad es constante V= 10 m/s, entonces no hay aceleración, se conserva el
movimiento rectilíneo uniforme. No hay ninguna fuerza que haga cambiar ese movimiento. (Ley
de la inercia)
2do tramo 30 – 50 seg
𝑎=
𝑣𝑓 − 𝑣𝑖
30𝑚/𝑠 − 10𝑚/𝑠 20𝑚/𝑠
=
=
= 1𝑚/𝑠 2
𝑡
20𝑠𝑒𝑔
20𝑠𝑒𝑔
𝐹 = 𝑚. 𝑎 = 600𝐾𝑔(1𝑚/𝑠 2 ) = 600𝑁
3er tramo 50 -65 seg
𝑣𝑓 − 𝑣𝑖
0𝑚/𝑠 − 30𝑚/𝑠 −30𝑚/𝑠
=
=
= −2𝑚/𝑠 2
𝑡
15𝑠𝑒𝑔
15𝑠𝑒𝑔
2𝑚
𝐹 = 𝑚. 𝑎 = 600𝐾𝑔 (− 2 ) = −1200𝑁
𝑠
𝑎=
12. F= 3200N
vi=95km/h = 26.3m/s
t=?
𝑎=
𝐹
3200𝑁
=
= −1.28𝑚/𝑠 2 𝑑𝑒𝑠𝑎𝑐𝑒𝑙𝑒𝑟𝑎𝑛𝑑𝑜
𝑚 2500 𝑘𝑔
𝑥=
𝑣𝑓 2 − 𝑣𝑖 2
0 − (26.3𝑚/𝑠)2
=
= 270.1𝑚
2𝑎
2(−1.28𝑚/𝑠 2 )
2𝑥
𝑎
𝑡=√
2(270.1𝑚)
= √
= 20.5𝑠𝑒𝑔
1.28𝑚/𝑠2
m=2500Kg
x=?
vf=0Km/h